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

asked Apr 9, 2023 125k views

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Mahmoud Haj Ali · Answer 1 · 2023-04-12T14:11:31+0000

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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 :

$\qquad \sf  \dashrightarrow \: \cfrac{12}{6.022 * 10 {}^(23) } \: \: g$

[ since 1 mole = 6.022 × 10²³ ]

$\qquad \sf  \approx2 * 10 {}^( - 23) \: \: g$

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 :

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

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

Question 4

The relationship between Molarity and molality can be expressed as :

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

Terms :

M = Molarity = ?

m = molality = 3 molal

d = density = 1.110 g/l

F = formula weight/molar mass = 40 g

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

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

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

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

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