Question

A student needs to dilute a 0.49 M Pb(NO3)2 solution to make 87.0 mL of 0.21 M Pb(NO3)2 . Set up the calculation by placing the values with the correct units into the equation. Then, calculate the volume, in milliliters, of the 0.49 M Pb(NO3)2 solution that is needed.

Answer 1

The initial volume of the solution of Pb(NO3)2 is 37.29 milliliters required to dilute for mentioned condition.

The volume will be calculated by the formula -

$M_{1}$$V_{1}$ = $M_{2}$$V_{2}$, where we have initial concentration and volume on Left Hand Side and final concentration and volume on Right Hand Side. Now, keeping the values in formula to find the final volume in milliliters.

0.49 × $V_{1}$ = 0.21 × 87

Performing multiplication on Right Hand Side of the equation

0.49 × $V_{1}$ = 18.27

Shifting the numerical value from Left Hand Side to Right Hand Side of the equation

$V_{1}$ = 18.27 ÷ 0.49

Performing division on Right Hand Side of the equation

$V_{1}$ = 37.29 millilitres

Hence, the volume is 37.29 milliliters.

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