How many moles of each substance are needed to prepare the follow...

Question

How many moles of each substance are needed to prepare the following solutions?a. 50.0 mL of 8.0% (m/v) KCl (MW=74.55g/mol)b. 200.0 mL of 7.5% (m/v) acetic acid (MW=60.05g/mol)

Accepted Answer

Welcome back, everyone. We need to calculate the amount of solute in moles needed to prepare a mL solution of 2.6% mass by volume copper sulfate with a molecular weight of 159. g per mole. Let's begin by recalling that. Our calculation for mass by volume per percent is found by taking or setting this equal to the following quotient where we have our mass of our solute divided by in our denominator, our volume of our solution. This is placed in parentheses and then multiplied by 100%. So we need to recognize that our given mass by volume percent for copper sulfate of 2.6% can be interpreted as 2.6 g of our solute copper sulfate divided by 100 ml of solution. So with this understanding, let's go into our formula to solve for the mass of copper sulfate first. So we would have on the left hand side of our equation, 2.6 g of our solute C US 04 divided by our volume of our solution. So this is going to be in terms of per 100 mL of solution. And then we're going to place this in parentheses and multiply it by 100. Since this is a mass by a volume percent, this is set equal to the other, the right hand side of our equation where we have our mass of our solute given or rather what we are solving for here. So we'll just say mass of C US divided by our volume of our solution given in the prompt as 500.0 mL of solution. Again, this is placed in parentheses and multiplied by 100 on the right hand side, since this also represents a mass by volume percent. So we need to isolate for mass, meaning we're going to multiply both sides by our denominator of 500. millilitres on the right hand side so that it can cancel out and we're going to divide both sides by 100. So that will cancel also 100 on the right. And we want to do the same thing on the left multiply it by 500. millis divided by 100. So that our 100 here cancels out and we will have milliliters being multiplied by grams in the numerator, meaning that we can cancel out milliliters from our denominator as well. And this will leave us with our unit as gramps, which is what we want for mass. So in our next line, this will simplify so that we have 1300 g of copper sulfate per or divided by 100 rather. And this is set equal to our mass of copper sulfate on the right hand side. So this quotient is going to simplify so that we find that our mass is equal to a value of 13 g of copper sulfate as our mass. So now that we have our massive copper sulfate in order to get to moles of our solute copper sulfate. We're going to begin with our mask that we found 13 g. And we're going to multiply by a conversion factor using sto geometry to go from grams of copper sulfate in the denominator to moles of copper sulfate in the numerator. And so here we can incorporate our given molar mass of copper sulfate with a molecular weight of 159.60 g of copper sulfate per mole of copper sulfate. So we have one mole in our new meter. So now to or canceling out our unit of grams, we're left with moles of copper sulfate as our final units. And this will give us our final answer equal to a value of 0.81454 g or moles. And we need to round this answer based on our given values and the prompt to the least precise measurement as two significant figures. Meaning our final answer when we move our decimal 20.2 decimal places will be 8.1 times 10 to the negative second power moles of copper sulfate. And this will be our final answer to complete this example as our moles of our solute. So I hope that this made sense and let us know if you have any questions.

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How many moles of each substance are needed to prepare the following solutions?a. 50.0 mL of 8.0% (m/v) KCl (MW=74.55g/mol)b. 200.0 mL of 7.5% (m/v) acetic acid (MW=60.05g/mol)

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