Question

1. what will be the mass of 1 atom of C-12 in grams?
2. What is the difference b/w molality&molarity.
3. Hydrogen gas is prepared in the laboratory by reacting dilute HCl with granulated zinc.
The following reaction takes place:

Zn + 2HCL ----› ZnCL2 +H2

Calculate the volume of hydrogen gas liberated at STP when 32.65g of zinc reacts with HCl. 1 mole of a gas occupies 22.7Litre volume at STP; Atomic mass= zn is 65.3u/amu.

4. the density of a 3 molal solution of NaOH is 1.110g m/l. calculate the molarity of the solution.​

Answer 1

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Here we go ~

Question 1

Mass of 1 mole C - 12 atom = 12 g

So, mass of 1 carbon - 12 atom = ( 12 / 1 mole ) g

that is :

$\qquad \sf  \dashrightarrow \: \cfrac{12}{6.022 \times 10 {}^{23} } \: \: g$

[ since 1 mole = 6.022 × 10²³ ]

$\qquad \sf  \approx2 \times 10 {}^{ - 23} \: \: g$

Question 2

Molarity :

Molarity is defined as " The number of moles of solute present in per litre of solution "

• Denoted as M = [ moles / litre ]

• change in temperature can cause change in Molarity, as the volume of solution varies with temperature.

• change in pressure can also cause change in Molarity, as volume is affected by pressure as well.

Molality :

Molality is defined as " Number of moles of solute present per kg mass of solvent "

• Denoted as m = [ moles / kg ]

• It isn't affected by any external factors like temperature or pressure, as mass of solvent is constant.

Question 3

As per the given reaction ~

$\qquad \sf  \dashrightarrow \: Zn + 2\:H Cl \rightarrow ZnCl_2 + H_2$

32.65 g of zinc reacted,

[ Number of moles of zinc reacted = mass of zinc reacted divided by its formula Weight ]

$\qquad \sf  \dashrightarrow \: number \: \: of \: \: moles = \cfrac{32.65}{65.3} \: \: mol$

$\qquad \sf  \dashrightarrow \: number \: \: of \: \: moles = \cfrac{1}{2} \: \: mol$

so, we can say that " half mole Zinc reacted with 1 mole of HCl to form half mole of Zinc chloride and half mole of Hydrogen gas "

And we already know that 1 mole of any gas occupies 22.7 litre volume at STP.

So, volume of Hydrogen gas Liberated :

$\qquad \sf  \dashrightarrow \: \cfrac{1}{2} \times 22.7$

$\qquad \sf  \dashrightarrow \: 11.35 \: \: litres$

Question 4

The relationship between Molarity and molality can be expressed as :

$\qquad \sf  \dashrightarrow \: M =\cfrac{ 1000(m \times d)}{1000+(m \times F)}$

Terms :

• M = Molarity = ?

• m = molality = 3 molal

• d = density = 1.110 g/l

• F = formula weight/molar mass = 40 g

$\qquad \sf  \dashrightarrow \: M =\cfrac{ 1000(3 \times 1.110)}{1000+(3 \times 40)}$

$\qquad \sf  \dashrightarrow \: M =\cfrac{ 1000( 3.330)}{1000+120}$

$\qquad \sf  \dashrightarrow \: M =\cfrac{ 3330}{1120}$

$\qquad \sf  \dashrightarrow \: M =2.973 \: mol \: l {}^{ - 1}$