63 g of a compound (Mol. Wt. = 126) was diss | vedclass

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

Question

(B) Step $1$: Calculate the number of moles of the solute. Moles $= \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$. S

Vedclass · Accepted Answer

$1.0$

Vedclass · Answer

(B) Step $1$: Calculate the number of moles of the solute. Moles $= \frac{	ext{Mass}}{	ext{Molar Mass}} = \frac{63 \ g}{126 \ g/mol} = 0.5 \ mol$. Step $2$: Calculate the total mass of the solution. Mass of solution $= 	ext{Mass of solute} + 	ext{Mass of solvent} = 63 \ g + 500 \ g = 563 \ g$. Step $3$: Calculate the volume of the solution using density. Volume $= \frac{	ext{Mass}}{	ext{Density}} = \frac{563 \ g}{1.126 \ g/mL} = 500 \ mL = 0.5 \ L$. Step $4$: Calculate the molarity. Molarity $(M) = \frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{0.5 \ mol}{0.5 \ L} = 1.0 \ M$.

$63 \ g$ of a compound (Mol. Wt. $= 126$) was dissolved in $500 \ g$ of distilled water. The density of the resultant solution is $1.126 \ g / mL$. The molarity of the solution is: (in $M$)

Similar Questions

For a solution containing $0.03659 \ g/mL$ of $HCl$ and a solution containing $0.04509 \ g/mL$ of acetic acid $(CH_3COOH)$,which of the following is true?

The molecular weight of glucose is $180$. $A$ solution of glucose which contains $18 \ g$ per litre is ........... $molal$.

If $5 \ g$ $NaOH$ is dissolved in $450 \ mL$ of solution,the molarity of the solution is: $(Na=23, O=16, H=1 \ g \ mol^{-1})$ (in $M$)

The density of a $1.0 \, m$ aqueous solution of glucose is $1.1 \, g/mL$. What is the molarity of the solution in $M$?

Which among the following is a unitless physical quantity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module