Question

calculate the molarity of mixed solution by adding 150ml of 0.1 M hcl to 200ml of 0.2 m hcl solution​

Answer 1

$\mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}}$

• Given :

=> Molarity of first solution (M1) = 0.1M

=> Volume of second solution (M2) = 0.2M

=> Volume of first solution (V1) = 150ml

=> Volume of second solution (V2) = 200ml

• To Find :

=> Molarity of mixed solution (M)

• Formula :

=> Molarity of mixure of two solution is given as...

$\: \: \: \: \: \: \sf{ \boxed{ \blue{ \bold{MV = M1V1 + M2V2}}}}$

=> V = volume of mixure = 150+200 = 350ml

• Calculation :

$\sf \: \implies \: M = \frac{M1V1 + M2V2}{V} \\ \\ \sf \implies \: M = \frac{(0.1 \times 150) + (0.2 \times 200)}{350} = 0.157M \\ \\ \\ \sf \: \: \bold{ \red{ \huge{\star}}} \: \: \boxed{ \boxed{ \huge{ \bold{ \blue{M = 0.16M}}}}} \: \: \bold{ \huge{ \red{ \star}}}$

Answer 2

Answer:

The molarity of the mixed solution will be 0.157 M.

Explanation:

Given:-

$M_{1}= 0.1 M$

$V_{1}= 150ml$

$M_{2}= 0.2 M$

$V_{2}= 200ml$

$Molarity= \frac{M_{1} V_{1} + M_{2} V_{2} }{V_{1} + V_{2} }$

$Molarity=\frac{55}{350}$

$Molarity= 0.157 M$

Therefore, the molarity of the mixed solution is 0.157M

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