Question

What is the initial volume, in milliliters, needed to prepare each of the following diluted solutions? (9.5)

a. $250mL$of $5.0\%\left(m/v\right)$glucose from $20.\%\left(m/v\right)$glucose

b. $45.0mL$of $1.0\%\left(m/v\right)Ca{\mathrm{CI}}_2$from $5.0\%\left(m/v\right){\mathrm{CaCI}}_2$

c.$100.mL$of$6.00M{H}_{2}S{O}_4$from$18.0M{H}_{2}S{O}_4$

Short answer

Part a. $250mL$of $5.0\%\left(m/v\right)$glucose from localid="1652689293387" $20.\%\left(m/v\right)$glucoselocalid="1653997382349" $=62.5mL$

Part b. $45.0mL$of $1.0\%\left(m/v\right)$${\mathrm{CaCI}}_2$from $5.0\%\left(m/v\right){\mathrm{CaCI}}_2$localid="1653997386546" $=9mL$

Part c. $100.mL$of localid="1652685928962" $6.00M{H}_{2}S{O}_4$from $18.0M{H}_{2}S{O}_4$localid="1653997392788" $=33mL$

Step-by-step solution

Introduction:

To make several of the preceding responses, they simply dilute the answers within the focus and microliters.

Given information :(Part a)

To prepare each of the following solutions, they needed dilute solutions in the initial volume and millilitres.

a. $250mL$of $5.0\%\left(m/v\right)$glucose from $20.\%\left(m/v\right)$glucose

b. $45.0mL$of $1.0\%\left(m/v\right)$${\mathrm{CaCI}}_2$from $5.0\%\left(m/v\right){\mathrm{CaCI}}_2$

c. $100mL$of localid="1653997404948" $6.00M{H}_{2}S{O}_4$fromlocalid="1653997408881" $18.0M{H}_{2}S{O}_4$

According to the dilution expression:

According to the dilution expression:

${\mathrm{C}}_1\times {\mathrm{V}}_1={\mathrm{C}}_2\times {\mathrm{V}}_2$

Here,

${\mathrm{C}}_1$= concentration of the concentrated solution

$V_1$= volume of the concentrated

${\mathrm{C}}_2$= concentration of the diluted solution

$V_2$ = volume of the diluted solution

Part (a) To calculate the glucose solution:

According to the problem,

${\mathrm{C}}_2=5.0\%\left(\mathrm{m}/\mathrm{v}\right)$

${\mathrm{V}}_2=250\mathrm{mL}$

${\mathrm{C}}_1=20\%\left(\mathrm{m}/\mathrm{v}\right)$

${\mathrm{V}}_1=?$

Rearrange the dilution expression to solve for the unknown quantity ${\mathrm{V}}_1$.

$\begin{array}{r}\begin{array}{r}{V}_1=\frac{C_2\times V_2}{C_1}\\ \end{array}\end{array}$

localid="1653997426274" $\begin{array}{r}\\ V_1=\frac{5.0\mathrm{\%}(\mathrm{m}/\mathrm{v})\times 250\mathrm{mL}}{20.0\mathrm{\%}(\mathrm{m}/\mathrm{v})}\end{array}$

$=62.5mL$

The required initial volume of glucose solution is$62.5mL$

Part (b)To calculate the CaCI2solution:

According to the problem,

${\mathrm{C}}_2=1.0\%\left(\mathrm{m}/\mathrm{v}\right)$

localid="1653997440212" ${\mathrm{V}}_2=45\mathrm{mL}$

${\mathrm{C}}_1=5.0\%\left(\mathrm{m}/\mathrm{v}\right)$

${\mathrm{V}}_1=?$

Rearrange the dilution expression to solve for the unknown quantity ${\mathrm{V}}_1.$

$\begin{array}{r}{V}_1=\frac{C_2\times V_2}{C_1}\end{array}$

localid="1653997450341" $\begin{array}{r}\\ V_1=\frac{1.0\mathrm{\%}(\mathrm{m}/\mathrm{v})\times 45\mathrm{mL}}{5.0\mathrm{\%}(\mathrm{m}/\mathrm{v})}\end{array}$

localid="1653997456441" $=9mL$

The required initial volume of${\mathrm{CaCI}}_2$solution is$9mL$.

Part (c) To calculate the H2SO4 solution:

According to the problem,

${\mathrm{C}}_2=6.0\mathrm{M}$

${\mathrm{V}}_2=100\mathrm{mL}$

${\mathrm{C}}_1=18.0\mathrm{M}$

${\mathrm{V}}_1=?$

Rearrange the dilution expression to solve for the unknown quantity ${\mathrm{V}}_1.$

$\begin{array}{r}{V}_1=\frac{C_2\times V_2}{C_1}\\ \end{array}$

localid="1653997477820" $\begin{array}{r}\\ V_1=\frac{6.0M\times 1000\mathrm{mL}}{18.0M}\end{array}$

localid="1653997481080" $=33mL$

The required initial volume of$H_{2}S{O}_4$solution islocalid="1653997485443" $33mL.$