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?

(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)#

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#