Rubbing alcohol is a 90% (by mass) solution of isopropyl alcohol,...

Question

Rubbing alcohol is a 90% (by mass) solution of isopropyl alcohol, C3H8O, in water. How many moles of isopropyl alcohol are in 50.0 g of rubbing alcohol? (LO 13.4)
(a) 45 mol

Accepted Answer

Hello everyone today, we have the following problem. Regular sugar syrup is a 20% by mass solution of sucrose in water, calculate the number of moles of sucrose and 300 g of regular sugar syrup. So the first thing I do is you want to account for our molar mass of sucrose and so we have the formula in the question. We have 12 carbons 22 hydrogen And 11 oxygen's. And what we're gonna do, we're gonna multiply by the molar mass found on the periodic table for carbon. That's 12.01 g per mole For hydrogen, that is 1.008 g per mole. And for oxygen that's going to be 16 g per mole. And so when we combine all of these Numbers, we're going to get a mass of .296 g per mole. And so we're gonna hang on to this number here. What we're going to do next is we're going to find our mass Of sucrose in this 300 g of regular syrup. And so we're gonna take 300 g of regular regular syrup. And we're gonna multiply by the 20% by mass that we have. So We can show that in the following way we have 20 g of our sucrose Per g of regular syrup When our units of regular syrup or grams of regular syrup canceled out. We're left with 60 g of sucrose. We're gonna go and hang on to that number for later. Lastly to find the moles of sucrose that we need. We're going to simply take our 60 g of sucrose And divide that by the Molar mass that we found, which is 342.296 g per mole. And so the way these units are set up is that our units of grams are going to cancel out, leaving us with 0.18 moles of sucrose as our final answer. And this is actually going to correlate with answer choice. C. Overall, I hope this helped, and until next time.

Rubbing alcohol is a 90% (by mass) solution of isopropyl alcohol, C3H8O, in water. How many moles of isopropyl alcohol are in 50.0 g of rubbing alcohol? (LO 13.4) (a) 45 mol

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