Calculate the molarity of a solution, given that its volume is 820 mL and that it contains 7.4 g of ammonium chloride, NH, CI.

The correct answer is 0.17 M

Explanation:

We have the following data:

V= 820 ml x 1 L/1000 ml = 0.82 L

m = 7.4 g

Ammonium chloride has the chemical formulae NH₄Cl. So, we calculate the molecular weight (MM) of NH₄Cl from the molar mass of N, H and Cl as follows:

MM NH₄Cl = 14 g/mol + (4 x 1 g/mol ) + 35.4 g/mol = 53.4 g/mol

Now we have to calculate the number of moles of NH₄Cl we have in the solution. For this, we divide the mass (m) into the molecular weight of NH₄Cl (MM):

moles NH₄Cl = m/MM = 7.4 g/53.4 g/mol = 0.138 moles

The molarity of the solution is the number of moles of NH₄Cl in 1 liter of solution. So, we have to divide the number of moles of NH₄Cl into the volume in liters as follows:

M = number of moles NH₄Cl/V = 0.138 moles/0.82 L = 0.169 mol/L ≅ 0.17 M

Molarity= 0.169M

Explanation:

Applying n= CV

V=820ml= 0.82dm3

n= m/M = 7.4/53.5 = 0.138mol

0.138= C× 0.82

C= 0.169M

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