Question

Molarity and molality of a solution of a liquid (mol wt=50) in aqueous solution is 9 and 10 respectively. What is the density of solution?​

Answer 1

Step-by-step explanation:

Solving:

$\:$

Molarity

$\large \mathtt{ = \dfrac{mass \: of \: solute}{molar \: mass \: of \: solute} \times \dfrac{100}{volume\: of \: solution} }$

$\:$

$\large \: \: \: \: \: \: \: \: \: \: : ⇒ \: \mathtt{m = \dfrac{w}{ {m} '} \times \dfrac{1000}{v} }$

$\:$

$\large \: : ⇒ \: \mathtt{w = \dfrac{m \times m' \times v}{1000} } \: \rm \longrightarrow(1)$

$\:$

Molarity

$\large \mathtt{ = \dfrac{mass \: of \: solute}{molar \: mass \: of \: solute} \times \dfrac{100}{mass\: of \: solvent} }$

$\:$

$\large \: \: \: \: \: \: \: \: \: \: : ⇒ \: \mathtt{m = \dfrac{w}{ {m} '} \times \dfrac{1000}{w'} }$

$\:$

$\large \: : ⇒ \: \mathtt{w = \dfrac{m \times m' \times w'}{1000} } \: \rm \longrightarrow(2)$

$\:$

Mass of Solution = Mass of Solvent + Mass of Solute

$\:$

⇒ w" = w + w'

$\:$

⇒ vd = w + w'

$\:$

⇒ w' = vd - w ________ (3)

$\:$

Putting Values in equation

$\:$

$\large : ⇒ \: \mathtt{ \dfrac{m \times m' \times v}{1000} = \mathtt{ \dfrac{m \times m' \times (vd - w)}{1000} } }$

$\:$

⇒ M × v = mvd - mw

$\:$

$\large⇒ \mathtt{ M × v = mvd -m \left( \dfrac{mm'v}{1000} \right)}$

$\:$

$\large⇒ \mathtt{ M = m \left( d - \dfrac{m'M}{100} \right)}$

$\:$

$\large⇒ \mathtt{ d = M \left( \dfrac{1}{m} + \dfrac{m'}{1000} \right)}$

$\:$

Density

$\large \mathtt{ = molarity\left( \dfrac{1}{molality} + \dfrac{molecular \: weight}{1000} \right)}$

$\:$

$\large \: \: \: \: \: \: \: : ⇒ \mathtt{ d = 9 \left( \dfrac{1}{10} + \dfrac{50}{1000} \right)}$

$\:$

D = 1.35g/cc

Answer 2

Answer:

Molarity

\large \mathtt{ = \dfrac{mass \: of \: solute}{molar \: mass \: of \: solute} \times \dfrac{100}{volume\: of \: solution} }=

molarmassofsolute

massofsolute

×

volumeofsolution

100

\:

\large \: \: \: \: \: \: \: \: \: \: : ⇒ \: \mathtt{m = \dfrac{w}{ {m} '} \times \dfrac{1000}{v} }:⇒m=

m

′

w

×

v

1000

\:

\large \: : ⇒ \: \mathtt{w = \dfrac{m \times m' \times v}{1000} } \: \rm \longrightarrow(1):⇒w=

1000

m×m

′

×v

⟶(1)

\:

Molarity

\large \mathtt{ = \dfrac{mass \: of \: solute}{molar \: mass \: of \: solute} \times \dfrac{100}{mass\: of \: solvent} }=

molarmassofsolute

massofsolute

×

massofsolvent

100

\:

\large \: \: \: \: \: \: \: \: \: \: : ⇒ \: \mathtt{m = \dfrac{w}{ {m} '} \times \dfrac{1000}{w'} }:⇒m=

m

′

w

×

w

′

1000

\:

\large \: : ⇒ \: \mathtt{w = \dfrac{m \times m' \times w'}{1000} } \: \rm \longrightarrow(2):⇒w=

1000

m×m

′

×w

′

⟶(2)

\:

Mass of Solution = Mass of Solvent + Mass of Solute

\:

⇒ w" = w + w'

\:

⇒ vd = w + w'

\:

⇒ w' = vd - w ________ (3)

\:

Putting Values in equation

\:

\large : ⇒ \: \mathtt{ \dfrac{m \times m' \times v}{1000} = \mathtt{ \dfrac{m \times m' \times (vd - w)}{1000} } } :⇒

1000

m×m

′

×v

=

1000

m×m

′

×(vd−w)

\:

⇒ M × v = mvd - mw

\:

\large⇒ \mathtt{ M × v = mvd -m \left( \dfrac{mm'v}{1000} \right)}⇒M×v=mvd−m(

1000

mm

′

v

)

\:

\large⇒ \mathtt{ M = m \left( d - \dfrac{m'M}{100} \right)}⇒M=m(d−

100

m

′

M

)

\:

\large⇒ \mathtt{ d = M \left( \dfrac{1}{m} + \dfrac{m'}{1000} \right)}⇒d=M(

m

1

+

1000

m

′

)

\:

Density

\large \mathtt{ = molarity\left( \dfrac{1}{molality} + \dfrac{molecular \: weight}{1000} \right)}=molarity(

molality

1

+

1000

molecularweight

)

\:

\large \: \: \: \: \: \: \: : ⇒ \mathtt{ d = 9 \left( \dfrac{1}{10} + \dfrac{50}{1000} \right)}:⇒d=9(

10

1

+

1000

50

)

\:

D = 1.35g/cc