Question

What is the concentration of a solution in mol/L when 80 g of calcium carbonate, CaCO3, is dissolved in 2 L of solution? How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl? A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required?

          What is the concentration of a solution in mol/L when 80 g of calcium carbonate, CaCO3, is dissolved in 2 L of solution?

How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl?

A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required?

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What is the concentration of a solution in mol/L when 80 g of calcium carbonate, CaCO3, is dissolved in 2 L of solution?

How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl?

A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required?

Added by Jessica A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Supreeta N

04/30/2023

Step 1

First, we need to find the moles of calcium carbonate (CaCO3) in 80 g. The molar mass of CaCO3 is 40.08 (Ca) + 12.01 (C) + 3 * 16.00 (O) = 100.09 g/mol. So, the moles of CaCO3 are: moles = mass / molar mass = 80 g / 100.09 g/mol = 0.799 moles Show more…

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What is the concentration of a solution in mol/L when 80 g of calcium carbonate, CaCO3, is dissolved in 2 L of solution? How many liters of 0.25 M NaCl solution must be measured to obtain 0.1 mol of NaCl? A student needs to prepare 250 ml of 0.1 M of Cd(NO3)2 solution. How many grams of cadmium nitrate are required?

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Transcript

00:01 Hi, in this question we are given with three parts.

00:03 In the first part of the question we are given with mass of calcium carbonate that is equal to 80 grams, volume of solution is 2 liter.

00:10 We are asked to calculate the concentration of the solution in terms of mole per liter.

00:16 We can say concentration is expressed in terms of molarity and molarity is equal to mass of solute that is calcium carbonate here divided by molar mass of calcium carbonate into volume of solution in terms of liter.

00:42 Now we can substitute all the values here.

00:45 In this manner now we can eliminate the common unit gram from here and on solving we are getting the or we can say the concentration of the solution equals to 0 .4 mole per liter.

00:59 This is the answer of first part.

01:01 Now in second part of the question we are given with the concentration of nacl solution that is equal to 0 .25 molar.

01:10 The moles of nacl in it is equal to 0 .21 mole.

01:14 We are asked to calculate the volume of solution in terms of liter.

01:18 We can say concentration is expressed in terms of molarity and molarity is equal to moles over volume.

01:25 We need to calculate the volume.

01:27 It will be equal to mole over molarity.

01:31 We can substitute the values here.

01:33 In this manner now we can eliminate the common unit mole from here since for molar we can write mole per liter...

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