Question

Initially a beaker contains 225.0 ml of a 0.350 M MgSO4 solution. Then 175.0 ml of water are added to the beaker. Find the concentration of the final solution. What is the volume?

Answer 1

Final volume is 400 mL

The moles in MgSO4 is 0.00788 mL

The new concentration is 0.197

Answer 2

Answer : The concentration of final solution is, 0.197 M and the final volume of solution is, 400 ml

Solution :

First we have to calculate the final volume of solution.

Final volume of solution = Volume of $MgSO_4$ + Volume of water added

Final volume of solution = 225.0 ml + 175.0 ml = 400 ml

Now we have to calculate the concentration of final solution.

Formula used :

$C_1V_1=C_2V_2$

where,

$M_1$ = concentration of $MgSO_4$ solution = 0.350 M

$V_1$ = volume of $MgSO_4$ solution = 225.0 ml

$M_2$ = concentration of final solution = ?

$V_2$ = volume of final solution = 400 ml

Now put all the given values in the above formula, we get the concentration of final solution.

$(0.350M)\times (225.0ml)=C_2\times (400ml)$

$C_2=0.197M$

Therefore, the concentration of final solution is, 0.197 M