In a water sample,the concentration of F^- io | vedclass

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

Question

(A) The concentration in $ppm$ (parts per million) is defined as the mass of solute in $mg$ per liter of solution.Given,concentration = $10 \ ppm = 10

Vedclass · Accepted Answer

$10^{-3}$

Vedclass · Answer

(A) The concentration in $ppm$ (parts per million) is defined as the mass of solute in $mg$ per liter of solution.Given,concentration = $10 \ ppm = 10 \ mg/L$.We know that $1 \ mg = 10^{-3} \ g$.So,$10 \ mg = 10 	imes 10^{-3} \ g = 10^{-2} \ g$.Thus,the concentration is $10^{-2} \ g/L$.To convert this to $\% W/V$ (grams per $100 \ mL$):Concentration in $\% W/V = \frac{	ext{mass of solute in } g}{	ext{volume of solution in } mL} 	imes 100$.Concentration = $\frac{10^{-2} \ g}{1000 \ mL} 	imes 100 = \frac{10^{-2}}{10} = 10^{-3} \% W/V$.

In a water sample,the concentration of $F^-$ ions is $10 \ ppm$ by weight-volume. What is the concentration of $F^-$ in $\% W/V$ in that solution?

Similar Questions

Two solutions of a substance (non-electrolyte) are mixed in the following manner: $480 \ mL$ of $1.5 \ M$ first solution $+ 520 \ mL$ of $1.2 \ M$ second solution. What is the molarity of the final mixture $.......... \ M$.

Assertion $(A)$: Molality of a solution increases with temperature.
Reason $(R)$: Molality expression does not involve any volume term.

Molarity is:

$A$ solution of sodium sulphate contains $92 \ g$ of $Na^{+}$ ions per kilogram of water. The molality of $Na^{+}$ ions in that solution in $mol \ kg^{-1}$ is:

$A$ $40 \%$ by mass sucrose solution is heated till it becomes $50 \%$ by mass. Calculate the mass of water lost from $100 \ g$ of the solution. (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module