Question

$\huge{\underline{\underline{Questions:-}}}$

1.the molality of dilute aqua solution of 0.05 M NaOH is
options:-
a)0.05 m
b)18.5 m
c)55.55 m
d)0.1 m
_________________________________________

2. The number of molecules present in One drop(0.05ml) of benzene(density = 0.78 g/ml)is nearly equal to${N_a = 6 \times {10}^{23}}$
options:-
a)6 × 10²³
b)3 × 10²¹
c)3 ×.10²⁰
d)6 × 10¹⁹

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Answer 1

Solution - 1).

Given:

=> Molarity = 0.05 M

=> Volume of solution = 1 L

To Find:

=> Molality

Formula used:

$\sf{\implies Molality=\dfrac{No.\;of\;moles\;of\;solute}{Volume\;of\;solution}}$

Here,

No. of moles of solute = 0.05 M

Now, we have to find mass of solution.

Let density of water = 1g/cm³

Then, mass of water = 1Kg

Put the following values in the formula,

$\sf{\implies Molality=\dfrac{No.\;of\;moles\;of\;solute}{Volume\;of\;solution}}$

$\sf{\implies Molality=\dfrac{0.05}{1}}$

$\large{\boxed{\boxed{\sf{\implies Molality=0.05}}}}$

Solution - 2).

Given:

=> Volume = 0.05 ml

=> Density = 0.78 g/ml

To Find:

=> The number of molecules present in One drop.

So,

As we know,

$\sf{\implies Density = \dfrac{Mass}{Volume}}$

$\sf{\implies Mass=Density\times Volume}$

So, Mass of 1 drop of benzene = 0.05 × 0.78

                                                   = 0.039 g

$\sf{\implies Moles\;of\;benzene=\dfrac{0.039}{78}}$

$\sf{\implies 5\times 10^{-4}}$

It is given that 1 mole of benzene contains $\sf{6\times 10^{23}\;molecules.}$

$\sf{So,\;molecules\;present\;in\;one\;drop=5\times 10^{-4}\times 6\times 10^{23}}$

$\large{\boxed{\boxed{\sf{\implies Molecules\;present\;in\;one\;drop=3\times 10^{20}\;molecules}}}}$

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore (1)\:Molality=0.05\:m}}$

${\bold{\therefore (2)\:Molecules=3\times10^{20}\:m}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\bold{For \: first \: question \: solution : } \\ \\ \underline \bold{given : } \\ \implies Molarity = 0.05 \: m \\ \\ \implies Volume \: of \: solution = 1 \: l \\ \\ \underline \bold{To \: Find : } \\ \implies Molality = ? \\ \\ \bold{According \: to\: given \: question : } \\ \\ \bold{By \: using \: formula \: of \: Molality} \\ \implies Molality = \frac{No. \: of \: moles \: in \: solution}{Volume \: of \: solution} \\ \\ \implies Molality = \frac{0.05}{1} \: \: \: (Density \: of \: water = 1 \: g {cm}^{3}) \\ \\ \bold{\therefore Molality = 0.05 m}$

$\\\bold{For \: second\:question \: solution :} \\ \\ \underline \bold{Given : } \\ \implies Volume \: of \: solution = 0.05 \: ml \\ \\ \implies Density \: of \: benzene = 0.78 \: gml \\ \\ \underline \bold{To \: Find : } \\ \implies No. \: of \: molecules \: present \\ \: \: \: \: \: \: \: \: \: in \: one \: drop = ? \\ \\ \bold{According \: to \: given \: question : } \\ \\ \bold{1 \: Mole \: of \: benzene = 6 \times {10}^{3} } \\ \\ \bold{By \: using\: formula \: of \: Density} \\ \implies Density = \frac{Mass}{Volume} \\ \\ \implies Mass = 0.05 \times 0.78 \\ \\ \bold{\implies Mass = 0.039 \: g} \\ \\ \bold{For \: Moles \: of \: benzene} \\ { \implies Moles \: of \: benzene }= \frac{0.039}{78} \\ \\ \implies Moles \: of \: benzene = 5 \times {10}^{ - 4} \\ \\ \bold{For \: Molecule \: present \: in \: one \: drop :} \\ \implies Molecules = 5 \times {10}^{ - 4} \times 6 \times {10}^{23} \\ \\ \implies Molecules = 30 \times {10}^{ - 4 + 23} \\ \\ \implies Molecules = 30 \times {10}^{19} \\ \\ \bold{\implies Molecules = 3 \times {10}^{20} }$