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3 years ago

How many grams of sodium hydroxide are present in 247.0 mL of a 0.300 M NaOH solution?

Chemistry

1 answer:

mamaluj [8]3 years ago

5 0

The grams of NaOH that are present in 247.0 ml of a 0.300M NaOH solution is 2.964 grams

calculation

Step 1: find moles of NaOH

moles = molarity x volume in liters

volume in liters = 247.0 /1000 = 0.247 liters

molarity = 0.300 M = 0.300 mol/l

moles = 0.300 mol/l x 0.247 =0.0741 moles

step 2: find mass of NaOH

mass= moles x molar mass

The molar mass of NaOH = 23 +1 +16= 40 g/mol

mass=0.0741 moles x40 g/mol =2.964 grams

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Read 2 more answers

A compound is 2. 00% H by mass, 32. 7% S by mass, and 65. 3% O by mass. What is its empirical formula? The final step is to use

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The empirical formula of this compound is $H_2SO_4$

Given the following data:

Percentage of H = 2.00%

Percentage of S = 32.7%

Percentage of O = 65.3%

Scientific data:

Molar mass of hydrogen (H) = 1.0 g/mol.

Molar mass of sulfur (S) = 32 g/mol.

Molar mass of oxygen (O) = 16 g/mol.

To determine the empirical formula of this compound:

Note: We would assume that the mass of the compound is 100 grams.

Hence, the mass of its constituent elements are:

Mass of hydrogen (H) = 2.00 grams

Mass of sulfur (S) = 32.7 grams

Mass of oxygen (O) = 65.3 grams

Next, we would determine the number of moles of each element by using this formula:

$Number\;of\;moles = \frac{mass}{molar\;mass}$

For hydrogen (H):

$Number\;of\;moles = \frac{2.00}{1}$

Number of moles = 2.0 moles

For sulfur (S):

$Number\;of\;moles = \frac{32.7}{32}$

Number of moles = 1.0 moles

For oxygen (O):

$Number\;of\;moles = \frac{65.3}{16}$

Number of moles = 4.0 moles

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