Question

1. accurately weigh 1.4710 g of the primary standard k₂cr₂o₇, dissolve it, and quantitatively transfer it to a 250 ml volumetric flask. calculate the concentration of this standard solution (c(k₂cr₂o₇)) and c(1/6 k₂cr₂o₇).

Explanation:

Step1: Calculate molar mass of $K_2Cr_2O_7$

The molar mass of $K$ is $39.10\ g/mol$, $Cr$ is $52.00\ g/mol$ and $O$ is $16.00\ g/mol$. So molar mass of $K_2Cr_2O_7$, $M(K_2Cr_2O_7)=2\times39.10 + 2\times52.00+7\times16.00=294.2\ g/mol$.

Step2: Calculate $c(K_2Cr_2O_7)$

The number of moles of $K_2Cr_2O_7$, $n=\frac{m}{M}=\frac{1.4710\ g}{294.2\ g/mol}=0.005\ mol$. The volume $V = 250\ mL=0.25\ L$. According to the formula $c=\frac{n}{V}$, $c(K_2Cr_2O_7)=\frac{0.005\ mol}{0.25\ L}=0.02\ mol/L$.

Step3: Calculate $c(1/6K_2Cr_2O_7)$

The equivalent - relationship shows that the molar mass of $1/6K_2Cr_2O_7$ is $\frac{M(K_2Cr_2O_7)}{6}$. The number of equivalents of $1/6K_2Cr_2O_7$ in the given mass is $n'=\frac{1.4710\ g}{\frac{294.2\ g/mol}{6}} = 0.03\ mol$. Then $c(1/6K_2Cr_2O_7)=\frac{0.03\ mol}{0.25\ L}=0.12\ mol/L$.

Answer:

$c(K_2Cr_2O_7)=0.02\ mol/L$, $c(1/6K_2Cr_2O_7)=0.12\ mol/L$