Question

Describe how you would prepare 2.00 L of each of the following solutions.
a. 0.250 M $NaOH$from solid $NaOH$
b. 0.250 M $NaOH$from 1.00 M$NaOH$ stock solution
c. 0.100 M $K_{2}Cr{O}_4$from solid $K_{2}Cr{O}_4$
d. 0.100 M $K_{2}Cr{O}_4$from 1.75 M stock solution $K_{2}Cr{O}_4$

Short answer

1. 0.250 M $NaOH$from solid$NaOH$is $20g$.
2. 0.250 M $NaOH$from 1.00 M $NaOH$stock solution is role="math" localid="1650273806125" $0.500L$.
3. 0.100 M $K_{2}Cr{O}_4$from solid$K_{2}Cr{O}_4$is role="math" localid="1650273885234" $38.838g$.
4. 0.100 M $K_{2}Cr{O}_4$from 1.75 M $K_{2}Cr{O}_4$stock solution is $0.1142L$.

Step-by-step solution

Definition

A solution's molarity is a concept that refers to the amount of solute contained in one litre of the solution. It may be stated mathematically as follows :

$Molarity=\frac{molesofsolute}{volumeofsolutioninliter}$

Determination of 0.250 M NaOH from solid NaOH

a)

$Molarity=moles\times \frac{1000}{V\left(mL\right)}\left(or\right)moles\times \frac{1}{V\left(L\right)}$

$\therefore Moles=\left(M\right)\left(V\right)$

$$\begin{gathered} Mass=\left(Molarmass\right)\left(moles\right) \\ =\left(40\frac{g}{mol}\right)\left(0.5mol\right) \\ =20g \end{gathered}$$

So, weigh$20g$ of $NaOH$, dissolve it in water, and generate up to$2.0L$ of flash.

Determination of 1.00 M NaOH from stock solution

b)

Dilution formula $N_1{V}_1=N_2{V}_2$ For $NaOH$

$M_1{V}_1=M_2{V}_2$

Let’s consider, $M_1=1.00m$, $V_1=?$

$M_2=0.250m,V_2=2.00L$

Therefore, $\left(1.00\right)\left(V_1\right)=\left(0.250\right)\left(2.00\right)$

$V_1=0.500L$

Therefore to make0.250M solution, use$0.500L$ of 1.00m$NaOH$ solution and dilute it to 2.00 L.

Determination of 0.100 M K2CrO4 from solid K2CrO4

c)

Molar mass of $K_{2}Cr{O}_4=194.19\frac{g}{mol}$

$$\begin{gathered} Molarity=\frac{Mass}{Molarmass}\times \frac{1}{Volume\left(L\right)} \\ 0.100=\frac{Mass}{194.19}\times \frac{1}{2.00} \\ =\left(0.250\right)\left(2.00L\right) \end{gathered}$$

role="math" localid="1650272910105" $$\begin{gathered} Mass=\left(Molarmass\right)\left(moles\right) \\ =0.100\times 194.19\times 2.00 \\ =38.838g \end{gathered}$$

So, weigh$38.838g$ of $K_{2}Cr{O}_4$, dissolve it in distilled water, and generate up to$2.0L$ of flash.

Determination 0.100 M K2CrO4 from 1.75 M K2CrO4 stock solution

d)

Dilution formula $N_1{V}_1=N_2{V}_2$for $K_{2}Cr{O}_4$

$M_1{V}_1=M_2{V}_2$

Let’s consider, $M_1=1.75m$, $V_1=?$

$M_2=0.100m,V_2=2.00L$

Therefore, $\left(1.75\right)\left(V_1\right)=\left(0.100\right)\left(2.00\right)$

$V_1=0.1142L$

Therefore to make $0.100M$solution, use $0.1142L$of $1.75M{K}_{2}Cr{O}_4$solution and dilute it to 2.00 L.