0-789 g of crystalline barium hydroxide is dissolved in water- For the neutralisation of this solution- 20 mL of N4HNO3 is required- How many molecules of water are present in one g mole of this base- Ba - 137-4- O - 15- N - 14- H - 1-

Let the molecular formula be Ba(OH)_2· xH_2O Mol. mass of Ba(OH)_2· xH_2O = 137.4 + (2× 16) + (2× 1) + 18x = 171.4 + 18x Eq. mass of Ba(OH)_ ...

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Standard XII

Chemistry

Stoichiometric Calculations

0.789 g of cr...

Question

$0.789 g$ of crystalline barium hydroxide is dissolved in water. For the neutralisation of this solution, $20 m L$ of $\frac{N}{4} H N O_{3}$ is required. How many molecules of water are present in one $g$ mole of this base? $(B a = 137.4, O = 15, N = 14, H = 1)$ .

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Solution

Let the molecular formula be $B a (O H)_{2} \cdot x H_{2} O$
Mol. mass of $B a (O H)_{2} \cdot x H_{2} O = 137.4 + (2 \times 16) + (2 \times 1) + 18 x$
$= 171.4 + 18 x$
Eq. mass of $B a (O H)_{2} \cdot x H_{2} O = \frac{171.4 + 18 x}{2}$
$20 m L \frac{N}{4} H N O_{3} \equiv 20 m L \frac{N}{4} B a (O H)_{2} \cdot x H_{2} O$
Amount of $B a (O H)_{2} \cdot x H_{2} O = \frac{(171.4 + 18 x)}{2 \times 4} \times \frac{20}{1000}$
$= \frac{171.4 + 18 x}{400} g$
Amount of $B a (O H)_{2} \cdot x H_{2} O = 0.789 g$
Hence, $\frac{171.4 + 18 x}{400} = 0.789$
or $171.4 + 18 x = 0.789 \times 400$
$x = \frac{144.2}{18} = 8.01 = 8$ .
Thus, $8 g$ moles of water molecules are present in one $g$ mole of the base.

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0.789 g of crystalline barium hydroxide is dissolved in water. For the neutralisation of this solution, 20 mL of N4HNO3 is required. How many molecules of water are present in one g mole of this base? (Ba=137.4,O=15,N=14,H=1).

$0.789 g$ of crystalline barium hydroxide is dissolved in water. For the neutralisation of this solution, $20 m L$ of $\frac{N}{4} H N O_{3}$ is required. How many molecules of water are present in one $g$ mole of this base? $(B a = 137.4, O = 15, N = 14, H = 1)$ .