Math, asked by Miracle901, 7 hours ago

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?​

Answers

Answered by ItzBrainlyLords
0

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

Answered by manav8852
0

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

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