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3 years ago

What is the mass in kg of 32500 g bag of sand

Chemistry

1 answer:

Paraphin [41]3 years ago

5 0

Answer:

32.5kg

Explanation:

if 1kg = 1000g,

then xkg=32500g

to convert, divide by 1000 32500/1000=32.5kg

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An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

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

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ 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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