Some Basic Concepts of Chemistry
When the volume of the solution is doubled, ______ becomes exactly half.
Molarity of $$4\%$$ (w/v) solution of $$NaOH$$ is:
Hint: Molarity (M) is the amount of a substance in a certain volume of solution.
Mathematically, it can be written as-
$$M=$$ number of moles of solute $$(n) \div$$ volume of solution $$(L)$$
Step 1: Find moles of $$NaOH$$
A per cent $$w/v$$ solution is calculated with the following formula using the gram as the base measure of weight (w):
$$% w/v$$ = g of solute/$$100$$ mL of solution
Weight = $$4$$gm
$$\text { Moles }=\frac{4\mathrm{~g}}{40 \mathrm{~g} / \mathrm{mol}}$$
Step 2: Find Molarity
As we know,
$$M=$$ number of moles of solute $$(n) \div$$ volume of solution $$(L)$$
The molecular mass of $$\mathrm{NaOH}$$ is $$=40$$
So, Molarity $$=\frac{4}{40} \times \frac{1000}{100}=1 \mathrm{M}$$
So, The correct option is $$D$$.
When the volume of the solution is doubled, ______ becomes exactly half.
1%(w/v) of an aqueous solution of NaOH is isomolar with:
Formula of Molarity (in ml) is .................. $$\times 100/v$$
What will be the molarity of a solution which contains $$5.85\; g$$ of $$NaCl(s)$$ per $$500$$ ml?
Stock phosphoric acid solution is $$85$$% $$H_3PO_4$$ and has a specific gravity of $$1.70$$ . Hence, molarity of $$H_3PO_4$$ solution is:
The amount of $$NaOH$$ in $$750ml$$ of $$0.2M$$ solution (Molecular weight $$=40$$) is:
An antifreeze mixture consists of $$40$$% ethylene glycol $$(C_2H_6O_2)$$ by weight in aqueous solution. If the density of this solution is $$ 1.05$$ g/mL, what is the molar concentration?
A semi molar solution is the one which contains:
The amount of sugar $$(C_{12}H_{22}O_{11}$$) required to prepare $$2L$$ of its $$0.1 M$$ aqueous solution is:
Dissolving 120 gm of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.5 g/mL. The molarity of the solution is: