Q: A student is preparing a solution and needs to convert 42,000 g/mol to grams for 1.2 moles. Which calculation correctly determines the mass required?

$42,000 \ \mathrm{g/mol} \times 1.2 \ \mathrm{mol} = 50,400 \ \mathrm{g}$

Q: Given a molecular weight of $42,000 \ \mathrm{g/mol}$, how many grams are needed to make a $0.5 \ \mathrm{mol}$ solution?

$42,000 \ \mathrm{g/mol} \times 0.5 \ \mathrm{mol} = 21,000 \ \mathrm{g}$

Q: If you are given $92,700 \ \mathrm{g/mol}$ as the molar mass of a protein, how many grams are in $0.25 \ \mathrm{mol}$?

$92,700 \ \mathrm{g/mol} \times 0.25 \ \mathrm{mol} = 23,175 \ \mathrm{g}$

Q: A researcher needs to prepare $2.5 \times 10^4$ grams of a compound with a molar mass of $42,000 \ \mathrm{g/mol}$. How many moles does this represent?

$\frac{2.5 \times 10^4 \ \mathrm{g}}{42,000 \ \mathrm{g/mol}} = 0.595 \ \mathrm{mol}$

Q: If a protein has a molar mass of $92,700 \ \mathrm{g/mol}$ and you have $18.54 \ \mathrm{g}$, how many moles of protein do you have?

$\frac{18.54 \ \mathrm{g}}{92,700 \ \mathrm{g/mol}} = 2.00 \times 10^{-4} \ \mathrm{mol}$

Q: If you need to prepare $1.34 \ \mathrm{mol}$ of a compound with a molar mass of $42,000 \ \mathrm{g/mol}$, how many grams do you need?

$1.34 \ \mathrm{mol} \times 42,000 \ \mathrm{g/mol} = 56,280 \ \mathrm{g}$

Q: A laboratory protocol requires $0.7 \ \mathrm{mol}$ of a protein with a molar mass of $92,700 \ \mathrm{g/mol}$. How many grams of protein are needed?

$0.7 \ \mathrm{mol} \times 92,700 \ \mathrm{g/mol} = 64,890 \ \mathrm{g}$

Q: A student mistakenly uses the formula $\text{grams} = \frac{\text{moles}}{\text{molar mass}}$ instead of $\text{grams} = \text{moles} \times \text{molar mass}$. What is the likely result if they try to calculate the mass of $2 \ \mathrm{mol}$ of a $42,000 \ \mathrm{g/mol}$ compound?

$\frac{2}{42,000} = 4.76 \times 10^{-5} \ \mathrm{g}$, which is much too small

Question 1

A student is preparing a solution and needs to convert 42,000 g/mol to grams for 1.2 moles. Which calculation correctly determines the mass required?

Accepted Answer

$42,000 \ \mathrm{g/mol} 	imes 1.2 \ \mathrm{mol} = 50,400 \ \mathrm{g}$

Question 2

If you have 92.7 kDa of a protein and want to know how many grams this is per mole, which of the following is correct? (1 Da = 1 g/mol)

Accepted Answer

$92,700 \ \mathrm{g/mol}$

Question 3

A calculation yields a value of 1.34. If this represents the concentration in mol/L, which of the following best describes the solution?

Accepted Answer

The solution has a concentration of $1.34 \ \mathrm{mol/L}$.

Question 4

Given a molecular weight of $42,000 \ \mathrm{g/mol}$, how many grams are needed to make a $0.5 \ \mathrm{mol}$ solution?

Accepted Answer

$42,000 \ \mathrm{g/mol} 	imes 0.5 \ \mathrm{mol} = 21,000 \ \mathrm{g}$

Question 5

If you are given $92,700 \ \mathrm{g/mol}$ as the molar mass of a protein, how many grams are in $0.25 \ \mathrm{mol}$?

Accepted Answer

$92,700 \ \mathrm{g/mol} 	imes 0.25 \ \mathrm{mol} = 23,175 \ \mathrm{g}$

Question 6

A researcher needs to prepare $2.5 	imes 10^4$ grams of a compound with a molar mass of $42,000 \ \mathrm{g/mol}$. How many moles does this represent?

Accepted Answer

$\frac{2.5 	imes 10^4 \ \mathrm{g}}{42,000 \ \mathrm{g/mol}} = 0.595 \ \mathrm{mol}$

Question 7

If a protein has a molar mass of $92,700 \ \mathrm{g/mol}$ and you have $18.54 \ \mathrm{g}$, how many moles of protein do you have?

Accepted Answer

$\frac{18.54 \ \mathrm{g}}{92,700 \ \mathrm{g/mol}} = 2.00 	imes 10^{-4} \ \mathrm{mol}$

Question 8

A solution is prepared by dissolving $25,000 \ \mathrm{g}$ of a compound with a molar mass of $42,000 \ \mathrm{g/mol}$ in enough water to make $1.0 \ \mathrm{L}$ of solution. What is the molarity of the solution?

Accepted Answer

$\frac{25,000 \ \mathrm{g}}{42,000 \ \mathrm{g/mol}} = 0.595 \ \mathrm{mol}$, so $0.595 \ \mathrm{M}$

Question 9

If you need to prepare $1.34 \ \mathrm{mol}$ of a compound with a molar mass of $42,000 \ \mathrm{g/mol}$, how many grams do you need?

Accepted Answer

$1.34 \ \mathrm{mol} 	imes 42,000 \ \mathrm{g/mol} = 56,280 \ \mathrm{g}$

Question 10

A laboratory protocol requires $0.7 \ \mathrm{mol}$ of a protein with a molar mass of $92,700 \ \mathrm{g/mol}$. How many grams of protein are needed?

Accepted Answer

$0.7 \ \mathrm{mol} 	imes 92,700 \ \mathrm{g/mol} = 64,890 \ \mathrm{g}$

Question 11

A student mistakenly uses the formula $	ext{grams} = \frac{	ext{moles}}{	ext{molar mass}}$ instead of $	ext{grams} = 	ext{moles} 	imes 	ext{molar mass}$. What is the likely result if they try to calculate the mass of $2 \ \mathrm{mol}$ of a $42,000 \ \mathrm{g/mol}$ compound?

Accepted Answer

$\frac{2}{42,000} = 4.76 	imes 10^{-5} \ \mathrm{g}$, which is much too small

Question 12

If a protein solution is labeled as $1.34 \ \mathrm{mol/L}$, how many grams of a $92,700 \ \mathrm{g/mol}$ protein are present in $500 \ \mathrm{mL}$?

Accepted Answer

$1.34 \ \mathrm{mol/L} 	imes 0.5 \ \mathrm{L} = 0.67 \ \mathrm{mol}$; $0.67 \ \mathrm{mol} 	imes 92,700 \ \mathrm{g/mol} = 62,109 \ \mathrm{g}$

Calculations Involving Molar Mass and Unit Conversions in Biochemistry

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