# A sample containing CH3NH3Cl was dissolved in water n then diluted to 500.00mL. A 15.00mL aliquot of the soln was titrated with 47.32mL of 0.1134M NaOH standard soln.Cal moles of CH3NH3Cl in d aliquot?mol of CH3NH3Cl in entire soln? molarity of d aliquot?

Feb 22, 2015

(a) $5.366 \times {10}^{- 3} m o l$

(b) $0.1788 m o l$

(c) $0.358 m o l . {l}^{- 1}$

$C {H}_{3} N {H}_{3} C l + N a O H \rightarrow C {H}_{3} N {H}_{2} + {H}_{2} O + N a C l$

This means that 1 mole of $C {H}_{3} N {H}_{3} C l$ reacts with 1 mole NaOH.

No. moles $N a O H = \frac{47.32 \times 0.1134}{1000} = 5.366 \times {10}^{- 3}$

So no. moles $C {H}_{3} N {H}_{2} C l = 5.366 \times {10}^{- 3}$ in 15ml

So no. moles in 500ml $= \frac{5.66 \times {10}^{- 3}}{15} \times 500$

$= 0.1788 m o l$

Molarity of aliquot$= \frac{n}{v} = \frac{5.366 \times {10}^{- 3}}{15 \times {10}^{- 3}} = 0.358 m o l . {l}^{- 1}$

Feb 22, 2015

So, you're dealing with a sample of $C {H}_{3} N {H}_{3} C l$ that you've diluted to a total volume of $\text{500.00 mL}$. The aliquot, which measures $\text{15.00 mL}$, is titrated with $\text{47.32 mL}$ of $N a O H$.

This is the key to solving the problem - you can determine the moles of $N a O H$, which will give you the moles of $C {H}_{3} N {H}_{3} C l$ in the aliquot, which in turn will get you to the number of moles in the $\text{500.00-mL}$ solution.

The balanced chemical equation for the reaction will be (I won't go into the net ionic equation)

$C {H}_{3} N {H}_{3} C {l}_{\left(a q\right)} + N a O {H}_{\left(a q\right)} \to N a C {l}_{\left(a q\right)} + C {H}_{3} N {H}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

The important thing here is the mole ratio between $C {H}_{3} N {H}_{3} C l$ and $N a O H$: 1 mole of the former will react with 1 mole of the latter. The number of $N a O H$ moles is

$C = \frac{n}{V} \implies {n}_{N a O H} = C \cdot V = \text{0.1134 M" * 47.32*10^(-3)"L}$

${n}_{N a O H} = \text{0.005366 moles NaOH}$

Automatically,

${n}_{C {H}_{3} N {H}_{3} C l} = \text{0.005366 moles}$

Use the volume of the aliquot to determine the molarity of the aliquot

C_("aliquot") = n/V = "0.005366 moles"/(15.00 * 10^(-3)"L") = "0.3577 M"

The number of moles in the $\text{500.00-mL}$ sample will be

$\text{500.00 mL" * ("0.005366 moles")/("15.00 mL") = "0.1789 moles}$ $C {H}_{3} N {H}_{3} C l$