If 3.00 liters of 2M HCl neutralizes 1.00 liter of NaOH, what is the molarity of the NaOH?

Chemistry

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Answer:

$M_{base}=6M$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this is a question about titration, which base solved by knowing that the HCl reacts with the NaOH in a 1:1 mole ratio, and therefore, we can write the following:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the base, NaOH, as shown below:

$M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{3L*2M}{1.00L}\\\\M_{base}=6M$

Regards!

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A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi

faust18 [17]

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

Or,

$\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT$

where,

$\pi$ = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = $62.364\text{ L.torr }mol^{-1}K^{-1}$

T = temperature of the solution = $26^oC=[273+26]K=299K$

Putting values in above equation, we get:

$4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol$