A solution is prepared with 70.0 g of HNO₃ and 130.0 g of H₂O. Th...

Question

A solution is prepared with 70.0 g of HNO₃ and 130.0 g of H₂O. The HNO₃ solution has a density of 1.21 g/mL. (9.4)

d. What is the molarity (M) of the solution?

Accepted Answer

Welcome back everyone. Vinyl acetate C 4h 602 is commonly used in producing industrial chemicals. A 20 g sample of vinyl acetate is dissolved in 70 g of water. The solution has obtained a density of 700.9 82 g per milliliter and determine the molar of the solution. Let's begin by recalling that molar is in terms of moles of our solute, which is our substance being dissolved divided by the volume of our solution and leaders. So in this case, we're going to need to begin by determining our moles of our solute vinyl acetate C 4h 602. We're going to need to take its mass given as 20 g. And we're going to need to multiply by a conversion factor to cancel our unit of grams in the denominator so that we can get some moles in the numerator. So this is where we're going to need to recall our molar mass of vinyl acetate from the periodic table where we have a mass of 86.90 g per mole of vinyl acetate. So this will allow us to then cancel our unit of grams leaving us with moles and we would result in our moles of our so equal to a value of 0. moles of vinyl acetate. So now with moles of our solute, we're next going to need to recall our formula for density where row is related to our mass of our substance divided by the volume of our solution. I'm sorry, that's our mass of our solution divided by its volume. So in this case, we're going to need our mass of our solution. Next, in which we're going to take our 20 g of our solute vinyl acetate added to our 70 g of our solvent water, which will give us a mass of our solution of 90 g. So now with our mass of our solution, we're next going to need to determine our denominator from our molar formula for volume. So we're going to need to rearrange our density formula. So moving for more space, we're going to multiply both sides by volume so that it cancels out on the right that will in our next line, give us row times volume equal to mass in which to cancel out row, we're going to divide both sides by row that will cancel out row on the left. And now we will have in our final line that our volume is equal to the mass divided by the density of our solution. So in this case, plugging in our mass of our solution, we found 90 g and we would divide by the density of our solution given in our prompt as 0. grams per milliliter. And recall that in our density formula volume is in units of liters. So in our denominator, we're going to multiply to go from milliliters in the numerator to leaders in the denominator. Recall that our prefix Milly tells us we have an equivalent of 10 to the negative third of our base unit leader in our denominator. So canceling out milliliters, we're left with grams per liter in our denominator in which we can then cancel out grams with grams in the numerator that leaves us with leader as our final unit. And for our volume of our solution, we'll find that we have a volume equal to 0. liters of our solution. So now with our volume of our solution and moles of our solute, we can go into calculating the molar of our solution by taking our moles of our solute above which we determined was 0. moles of vinyl acetate C 4h 602. This is then divided by our volume of our solution which we just determined above as 0.9165 liters of solution. So in our calculators, this quotient will simplify to our molar of our solution equal to a value of about two, about 2.53 molar. And so 2.53 molar will be our final answer to determine the molar of our solution of vinyl acetate in water. I hope that this made sense and let us know if you have any questions.

A solution is prepared with 70.0 g of HNO₃ and 130.0 g of H₂O. The HNO₃ solution has a density of 1.21 g/mL. (9.4) d. What is the molarity (M) of the solution?

Concetti chiave

Molarity

Calculating Moles

Volume of Solution

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