Question

What mass (in grams) of Mg(NO3)2 is present in 151 mL of a 0.350 M solution of Mg(NO3)2?

A) 7.84 g

B) 0.0529 g

C) 4.56 g

D) 64.0 g

Answer 1

Answer is: A) 7.84 g.

V(Mg(NO₃)₂) = 151 mL ÷ 1000 mL/L.

V(Mg(NO₃)₂) = 0.151 L; volume of the magnesium nitrate.

c(Mg(NO₃)₂) = 0.352 M; molarity of the solution.

n(Mg(NO₃)₂) = V(Mg(NO₃)₂) · c(Mg(NO₃)₂).

n(Mg(NO₃)₂) ) = 0.151 L · 0.352 mol/L.

n(Mg(NO₃)₂) = 0.0531 mol; amount of the substance.

M(Mg(NO₃)₂) = Ar(Mg) + 2Ar(N) + 6Ar(O) · g/mol.

M(Mg(NO₃)₂) = 24.3 + 2·14 + 6·16 · g/mol.

M(Mg(NO₃)₂) = 148.3 g/mol; molar mass.

m(Mg(NO₃)₂) = n(Mg(NO₃)₂) · M(Mg(NO₃)₂).

m(Mg(NO₃)₂) = 0.0531 mol · 148.3 g/mol.

m(Mg(NO₃)₂) = 7.84; mass of magnesium nitrate.

Answer 2

The correct option is$\boxed{{\mathbf{A}}{\mathbf{. 7}}{\mathbf{.84 g}}}$.

Further explanation:

Mole is the S.I. unit. The number of moles is calculated as the ratio of mass of the compound to that of molar mass of the compound.

Molar mass also known as molecular weight is the sum of the total mass in grams of all the atoms that make up a mole of a particular molecule that is the mass of 1 mole of a compound. Its S.I unit is g/mol.

The expression to relate number of moles, mass and molar mass of compound is as follows:

${\text{Number of moles}}=\frac{{{\text{mass of the compound}}}}{{{\text{molar mass of the compound}}}}$       …… (1)

Molarity:

The molarity is the concentration of the solution and is equal to the number of moles of the solute dissolved in liter of the solution.

The expression to relate molarity (M), volume (V), and the number of moles (n) is as follows:

${\text{M}}=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}$                 …… (2)

Here, V is a volume of solution in liters and n is a number of moles of solute.

The conversion factor to convert volume in liter (L) to (mL) is written as follows:

${\text{1L}}=1000\;{\text{mL}}$

Given volume of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$solution is 151 mL.

Therefore, 151 mL of volume is converted into (L) as follows:

$\begin{aligned}{\text{Volume}}&=\,151{\text{ mL}}\left({\frac{{{\text{1}}\;{\text{L}}}}{{1000\;{\text{mL}}}}}\right)\\&=0.151{\text{ L}}\\\end{aligned}$

On rearranging equation (2) for n, we get,

${\text{n(mol)}}={\text{M}}\times{\text{V(L)}}$                   …… (3)

Concentration of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$ solution (M) is 0.350 M.

Volume of${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$ solution is 0.151 L.

Substitute these values in equation (3) to calculate the number of moles (n).

$\begin{aligned}{\text{n}}\left({{\text{mol}}}\right)&={\text{(0}}{\text{.350 M)}}\left( {\frac{{1\;{\text{mol/L}}}}{{{\text{1}}\;{\text{M}}}}}\right)\times\left({{\text{0}}{\text{.151 L}}}\right)\\&={\text{0}}{\text{.05285 mol}}\\\end{aligned}$

The number of moles (n) of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$is 0.05285 mol.

The formula to calculate the mass of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$.

${\text{Mass of Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}={\text{moles of Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\times {\text{molar mass of}}\;{\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$         …… (4)

Molar mass of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$is 148.3g/mol.

Moles of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$is 0.05285 mol.

Substitute these values in equation (4) to calculate the mass of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$.

$\begin{aligned}{\text{Mass of Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}&=\left({{\text{0}}{\text{.05285}}\;{\text{mol}}}\right)\times\left({{\text{148}}{\text{.3}}\;{\text{g/mol}}}\right)\\&={\text{7}}{\text{.8376 g}}\\&\approx{\text{7}}{\text{.84 g}}\\\end{aligned}$

Hence, 7.84 g of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$is present in 151 mL of 0.350 M of ${\text{Mg(N}}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}$solution.

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Answer details:

Grade: SeniorSchool

Subject: Chemistry

Chapter: Solutions

Keywords: mole, molarity, volume, concentration, 151 mL,0.350M, 0.05285 mol, 148.3g/mol.