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3 years ago

Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius

Chemistry

1 answer:

kotykmax [81]3 years ago

8 0

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

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To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

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$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

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