Question

A 250 mL glucose solution is prepared by dissolving 15 g of glucose in water. What is the mass by volume percentage of glucose in solution? option A 5.66% option B 6% option C 94% option D 94.34%​

Answer 1

Answer:

• Given weight of solute = 15 g

• Given mass of solution = 250 ml

• % W/V = ?

Mass by volume percentage of glucose :

$\implies \rm \% \dfrac{W}{V} = \dfrac{W_{(solute)} }{Volume_{(solution)}} \times 100$

$\implies \rm \% \dfrac{W}{V} = \dfrac{15 }{250} \times 100$

$\implies \rm \% \dfrac{W}{V} = \dfrac{15 }{25} \times 10$

$\implies \rm \% \dfrac{W}{V} = \dfrac{3 }{5} \times 10$

$\implies \rm \% \dfrac{W}{V} = \dfrac{3 0}{5}$

$\implies \underline{\boxed{\red{\bf \% \dfrac{W}{V} = 6 \: \% }}}$

Hence, option B) 6% is the correct answer.

Concentration terms:

$\bullet \:\sf Molarity = \dfrac{Moles_{(solute)}}{Volume_{(Solution)}}$

$\bullet\: \sf Molality = \dfrac{Moles_{(solute)}}{Weight_{(Solvent)}}$

$\bullet\: \sf \chi_{solvent}= \dfrac{Moles_{(solvent)}}{Moles_{(Solvent)} + Moles_{(Solute)}}$

Answer 2

$\sf\huge\bold{Answer:}$

$\fbox{Option B is correct.}$

Given :

• Total volume of solution = 250 mL

• Given mass of glucose = 15 g

To find :

mass by volume percentage (m/v) of glucose in solution

Solution :

$\fbox{Formula used :}$

$\bf\red{ :⟶m/v \: \%\:of\:solute = \frac{mass \: of \: solute}{volume \: of \: solution} \times 100\%}$

• mass of solute(glucose) = 15g
• Volume of solution(glucose+water) = 250 mL (i.e 250 g). [1mL=1g]

Inserting values in formula, we get

$\\ \\ \\ \sf{ :⟶m/v \: \%\:of\:glucose = \frac{15}{250} \times 100\%}$

$\sf{:⟶m/v \: \%\:of\:glucose = }$$\displaystyle{\sf{ \cancel{ \frac{15}{250} }×100\% }}$

$\sf{ :⟶m/v \: \%\:of\:glucose = \frac{3}{50} \times 100\%}$

$\sf{:⟶m/v \: \%\:of\:glucose = }$$\displaystyle{\sf{ \cancel{ \frac{3×100}{50} }\% }}$

$\sf{ :⟶m/v \: \%\:of\:glucose = 3×2\%}$

$\bf{ :⟶m/v \: \%\:of\:glucose = 6\%}$

$\bf\huge\purple{mass\: by \: volume \: \%\:of\:glucose\: is \:}$

$\huge \implies \boxed{6\%}$