Question

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?

Answer 1

(a) `5.366xx10^(-3)mol`

(b) `0.1788mol`

(c) `0.358mol.l^(-1)`

`CH_3NH_3Cl+NaOHrarrCH_3NH_2+H_2O+NaCl`

This means that 1 mole of `CH_3NH_3Cl` reacts with 1 mole NaOH.

No. moles `NaOH = (47.32xx0.1134)/(1000)=5.366xx10^(-3)`

So no. moles `CH_3NH_2Cl=5.366xx10^(-3)` in 15ml

So no. moles in 500ml `=(5.66xx10^(-3))/(15)xx500`

`=0.1788mol`

Molarity of aliquot`=n/v=(5.366xx10^(-3))/(15xx10^(-3))=0.358mol.l^(-1)`

Answer 2

So, you're dealing with a sample of `CH_3NH_3Cl` that you've diluted to a total volume of `"500.00 mL"`. The aliquot, which measures `"15.00 mL"`, is titrated with `"47.32 mL"` of `NaOH`.

This is the key to solving the problem - you can determine the moles of `NaOH`, which will give you the moles of `CH_3NH_3Cl` in the aliquot, which in turn will get you to the number of moles in the `"500.00-mL"` solution.

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

`CH_3NH_3Cl_((aq)) + NaOH_((aq)) -> NaCl_((aq)) + CH_3NH_(2(aq)) + H_2O_((l))`

The important thing here is the mole ratio between `CH_3NH_3Cl` and `NaOH`: 1 mole of the former will react with 1 mole of the latter. The number of `NaOH` moles is

`C = n/V => n_(NaOH) = C * V = "0.1134 M" * 47.32*10^(-3)"L"`

`n_(NaOH) ="0.005366 moles NaOH"`

Automatically,

`n_(CH_3NH_3Cl) = "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 `"500.00-mL"` sample will be

`"500.00 mL" * ("0.005366 moles")/("15.00 mL") = "0.1789 moles"` `CH_3NH_3Cl`