Question

n/takeAssignment/takeCovalentActivity.do?locator assignment-takent.....Maps[References)How many grams of solute are present in each of the following aqueous solutions?(a) 575 mL of a 2.41 M solution of nitric acid, HNO39(b) 1.65 L of a 0.290 M solution of alanine, C3H7NO2Submit Answer9(c) 320.0 mL of a 0.00679 M solution of calcium sulfate, CaSO49Try Another Version 10 item attempts remaining

Answer 1

In this question, we are given the volume and the Molarity of a solution, from these two informations, we will be using the Molarity formula, the formula is:

M = n/V

This formula is used to find the number of moles (n)

M = 2.41

V = 575 mL, but we need it in Liters, 0.575 L

Now we add these values into the formula:

2.41 = n/0.575

n = 0.575 * 2.41

n = 1.38 moles of HNO3

Now that we have the number of moles of HNO3, we can find the mass in grams using this number of moles, 1.38 moles, and the molar mass of HNO3, which is 63.01g/mol

63.01g = 1 mol

x grams = 1.38 moles

x = 1.38 * 63.01

x = 86.95 grams of HNO3

In this solution we have 86.95 grams