Question

If 2.0×10−4 moles of S2O2−8 in 170 mL of solution is consumed in 170 seconds , what is the rate of consumption of S2O2−8?

Answer 1

Answer : The rate of consumption of $S_2O_2^{-8}$ is, $7.0\times 10^{-6}M/s$

Explanation : Given,

Moles of $S_2O_2^{-8}$ = $2.0\times 10^{-4}mol$

Volume of solution = 170 mL = 0.170 L     (1 L = 1000 mL)

Time = 170 s

First we have to calculate the concentration.

$\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}$

$\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}$

$\text{Concentration}=1.2\times 10^{-3}M$

Now we have to calculate the rate of consumption.

$\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}$

$\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}$

$\text{Rate of consumption}=7.0\times 10^{-6}M/s$

Thus, the rate of consumption of $S_2O_2^{-8}$ is, $7.0\times 10^{-6}M/s$