Two solutions labeled as 3,M,HCl and 1,M,HCl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $M_1 = 3\,M$ and $M_2 = 1\,M$.Case $1$: When mixed in the ratio $x : y$ by volume,the resultant molarity $M_R$ is given by:$M_R = \frac{M_1x

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(D) Given $M_1 = 3\,M$ and $M_2 = 1\,M$.Case $1$: When mixed in the ratio $x : y$ by volume,the resultant molarity $M_R$ is given by:$M_R = \frac{M_1x + M_2y}{x + y} = 1.5$$\frac{3x + y}{x + y} = 1.5$$3x + y = 1.5x + 1.5y$$1.5x = 0.5y \Rightarrow y = 3x$.Case $2$: When mixed in the ratio $y : x$ by volume,the new resultant molarity $M'_R$ is:$M'_R = \frac{M_1y + M_2x}{y + x}$Substituting $y = 3x$:$M'_R = \frac{3(3x) + 1(x)}{3x + x} = \frac{9x + x}{4x} = \frac{10x}{4x} = 2.5\,M$.

Two solutions labeled as $3\,M\,HCl$ and $1\,M\,HCl$ are mixed in the ratio of $x : y$ by volume and the molarity of the resulting solution is $1.5\,M$. The molarity of the resulting solution if they are mixed in the ratio $y : x$ by volume is ....... $M$.

Similar Questions

An antifreeze solution is prepared from $222.6 \ g$ of ethylene glycol $(C_2H_6O_2)$ and $200 \ g$ of water. Calculate the molality of the solution. If the density of the solution is $1.072 \ g \ mL^{-1}$,then what shall be the molarity of the solution?

Which of the following modes of expressing concentration is independent of temperature?

An aqueous solution contains $28\%$ by weight of a liquid $A$ $(m_w = 140)$. The mole fraction of liquid $A$ is:

If $80 \ g$ of $NaOH$ is mixed with $54 \ g$ of water to form a solution,calculate the mole fraction of water.

The molarity of the solution when $4.9 \ g$ of $H_2SO_4$ is dissolved in $250 \ mL$ of solution is: (in $M$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module