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bazaltina [42]
3 years ago
5

Label 0.15 m cacl2 as isotonic, hypotonic, or hypertonic in comparison to 0.9% nacl (0.15 m nacl).

Chemistry
1 answer:
zlopas [31]3 years ago
3 0
To know if the solutions are isotonic, hypertonic or hypotonic, we have to determine the osmotic pressure of each solution.

P = iMRT
where
i is the number of ions dissociated for strong electrolytes
M is the molarity
R is the universal gas constant
T is the absolute temperature

For 0.15 M CaCl₂, i = 3 (1 for Ca⁺ and 2 for Cl⁻).
P = (3)(0.15)RT= 0.45RT

For 0.15 NaCl, i = 2 (1 for Na⁺ and 1 for Cl⁻).
P = 2(0.15)RT = 0.3RT

Therefore, the P for CaCl₂ is GREATER than the P for NaCl. <em>Hence, the CaCl₂ solution is hypertonic.</em>
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Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
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According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

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Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

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Similarly, number of moles of oxygen will be:

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The ratio of number of moles of carbon and oxygen will be:

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The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


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