What mass of #"ammonium chloride"# solute is required to prepare a #200*mL# of solution for which the concentration was #0.100*mol*L^-1#?

3 Answers
Nov 23, 2017

#1.07g#

Explanation:

First calculate the mass of one mole of #NH_4Cl#. This would produce a #1.00 (mol)/L# solution in 1 Liter. Then take #1/10# of that to make #0.100# mol, and then #1/5# of that for the #200.00# mL solution (#200/1000#).

M.W. #NH_4Cl# = #14 + 4 + 35.5 = 53.5g#

#53.5 xx 1/10 xx 1/5 = 1.07g# in #200.00# mL solvent.

Nov 23, 2017

Approx. #1*g# solute are required....

Explanation:

To define the #"concentration"# we use the quotient....

#"concentration"="Moles of solute"/"Volume of solution"#

And thus........

#"moles of ammonium chloride"="volume"xx"concentration"#

#=200.0*cancel(mL)xx10^-3*cancelL*cancel(mL^-1)xx0.100*mol*cancel(L^-1)#

#=0.020*mol#

And we take the product of this molar quantity by the molar mass of ammonium chloride to get the required mass of solute....

#=0.020*molxx53.49*g*mol^-1=??*g#.

Nov 23, 2017

#"1.0698 g"#

Explanation:

Moles of solute#= "0.100 mol"/cancel"L" × 200.00 cancel"ml" × (10^-3 cancel"L")/cancel"ml"= "0.02 mol"#

Molar mass of #"NH"_4"Cl"# is #"53.491 g/mol"#

Mass of solute#= 0.02 cancel"mol" × "53.491 g"/cancel"mol" = "1.0698 g"#