Question

480 ml of 1 M solution is mixed with another 520 ml of 1.2 M solution. Then the molarity of resulting solution is …...….

Answer 1

AnswEr :

Resultant Molarity is 1.104 M

Given that,

• $\sf Molarity,M_1 = 1 M$

• $\sf Volume \ Occupied,L_1 = 480 ml$

• $\sf Molarity,M_2 = 1.2 M$

• $\sf Volume \ Occupied,L_2 = 520 ml$

Volume of the resulting solution would be : 420 + 580 = 1000ml = 1 L

We have to find the Molarity of Resulting Solution

WKT,

$\boxed{\boxed{\sf M_1V_1 + M_2V_2 = M_3V_3}}$

Substituting the values,

$\implies \: \bigg(\sf \: 1 \times \dfrac{480}{1000} \bigg) + \bigg(1.2 + \dfrac{520}{1000} \bigg) = M \times \dfrac{1000}{1000} \\ \\ \implies \: \sf \: (1 \times 0.48) + (1.2 \times 0.52) = M \\ \\ \implies \: \sf \: M = 0.624 + 0.48 \\ \\ \implies \: \underline{ \boxed{ \sf Molarity = 1.104 \: M \ }}$

Answer 2

Given values :-

$\color{gold} {\tt{\bigstar}}\ \displaystyle{\tt{Molarity\ of\ the\ first\ solution(M_1)=1\ M}}\\\color{gold}\bigstar\ {\displaystyle{\tt{Volume\ of\ the\ first\ solution(V_1)=480\ ml}}}\\\\\color{brown} {\tt{\blacksquare}}\ \displaystyle{\tt{Molarity\ of\ the\ second\ solution(M_2)=1.2\ M}}\\\color{brown} {\tt{\blacksquare}}\ \displaystyle{\tt{Volume\ of\ the\ second\ solution(V_2)=520\ ml}}$

To calculate :-

$\longmapsto \texttt{The molarity of the resulting solution}\tt{(M_3).}$

Answer :-

$\texttt{Volume of the resulting solution :}\\\tt{=480+520}\\\color{green} {\tt{=1000\ ml }}$

$\tt{We\ know,}\\\boxed{\color{green} {\displaystyle{\tt{M_1V_1+M_2V_2=M_3V_3}}}}$

$\underline{\tt{Putting\ the\ known\ values\ in\ the\ above\ equation:-}}$

$\tt{(1\times480)+(1.2\times520)=M_3\times1000}\\\\\tt{\dashrightarrow 480+624=1000M_3}\\\\\tt{\dashrightarrow 1104=1000M_3}\\\\\boxed{\boxed{\tt{\dashrightarrow M_3=1.104\ M}}}$

$\large \color{green} {\underline{\tt{So,\ the\ molarity\ of\ the\ resulting\ solution\ is\ 1.104\ M.}}}$