Question

Convert 0.169 Molar HC2H3O2 to % (m/V).

Answer 1

Answer:

We define

%

w/v

as:

%

w/v

=

solute mass in g

solution volume in mL

×

100

%

By unit conversion, we have that

g

×

mol

g

→

mol

, and that

mL

×

1 L

1000 mL

→

L

. Define:

m

s

o

l

u

t

e

for the solute mass in

g

V

s

o

ln

for the solution volume in

mL

M

s

o

l

u

t

e

for the molar mass of the solute in

g/mol

Therefore:

Molarity

=

(

m

s

o

l

u

t

e

V

s

o

ln

×

100

%

)

×

1000

mL/L

100

%

×

M

s

o

l

u

t

e

Or, rewriting in terms of

%

w/v

and implied unit cancellation, we have:

Molarity

(

mol

L

)

=

1000

M

s

o

l

u

t

e

%

w/v

100

%

As an example, if we have a

37

%

w/v

aqueous

HCl

solution, then:

Molarity

(

mol

L

)

=

1000

36.4609

37

%

w/v

100

%

=

10.15 mol/L

Answer 2

Answer:

Explanation:

Concept:

The number of moles of solute in one litre of solution is referred to as molarity. And the mass of the solute in grams contained in $100 \mathrm{~mL}$ of solution is $\mathrm{m} / \mathrm{v}$.

Given:

The value of Molarity is $0.169 \mathrm{M}$

The Molar mass value of $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ is $=((12 \times 2)+(1 \times 4)+(16 \times 2)=60 \mathrm{~g} / \mathrm{mol}$

Find:

Calculate the percent$(m/V)$ of $0.169$ Molar $\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$

Solution:

$0.169$ molar indicates there are $0.169$ moles of solute in $1 \mathrm{~L}$ of solution.

$0.169 M=\frac{0.169 \mathrm{~mol}}{1 L}$

Both the numerator and the denomination should be multiplied by ten.

$\frac{\frac{0.169 mol}{10} }{\frac{1L}{10} } =\frac{0.0169 mol}{0.1L}$$=\frac{0.0169 mol}{100 mL}$

Note :$1 \mathrm{~L}=1000 \mathrm{~mL}$ so $0.1 \mathrm{~L}$ will be $100 \mathrm{~mL}$.

That is, in $100 \mathrm{~mL}$, $0.0169 \mathrm{~mol}$ is present. Use the formula below to convert $0.0169 \mathrm{~mol}$ to gram.

$\begin{aligned}\text { number of moles }=& \frac{\text { given mass }}{\text { molar mass }} \\0.0169 \mathrm{~mol} &=\frac{\text { given mass }}{60 \mathrm{~g} / \mathrm{mol}} \\&=0.0169 \text { mol } \times 60 \mathrm{~g} / \text { mol } \\&=1.014 \mathrm{~g}\end{aligned}$

Here $1.014 \mathrm{~g}$ of solute is present in $100 \mathrm{~mL}$ of solution

Therefore $0.169 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ will be equal to $1.014 \%(\mathrm{~m} / \mathrm{v})$

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