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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

ntakeAssignmenttakeCovalentActivitydolocator assignmenttakentMapsReferencesHow many grams of solute are present in each of the following aqueous solutionsa 575 class=

Respuesta :

According to the explanation given in the last session, for letter B we will have to find the mass for this solution using the Molarity formula again:

M = n/V

0.290 = n/1.65

n = 0.290 * 1.65

n = 0.478 moles of C3H7NO2

The molar mass of C3H7NO2 is 89.09g/mol

89.09 grams = 1 mol

x grams = 0.478 moles

x = 0.478 * 89.09

x = 42.58 grams of C3H7NO2

In this solution, we have 42.58 grams of C3H7NO2

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