Question

How many grams of NaOH will be need to make250ml of 1.5M solution

Answer 1

Answer:

• 15 grams of NaOH will make a molarity of 1.5 M.

Given:

• Volume of solution (V) = 250 ml.
• Molarity of solution (M) = 1.5 M

Explanation:

$\rule{300}{1.5}$

From the Molarity formula we Know,

$\large \bigstar \; {\boxed{\tt M = \dfrac{n}{V_{(In \; ml)}} \times 1000}}$

$\bold{Here}\begin{cases}\text{n Denotes Number of moles} \\ \text{V Denotes Volume (In ml)} \\ \text{M Denotes Molarity}\end{cases}$

Now,

$\large {\boxed{\tt M = \dfrac{n}{V_{(In \; ml)}} \times 1000}}$

Substituting the Values,

$\longmapsto \large{\tt 1.5 \; M = \dfrac{n}{250 \; ml} \times 1000}$

$\longmapsto \large{\tt 1.5 = \dfrac{n}{250} \times 1000}$

$\longmapsto \large{\tt 1.5 = \dfrac{1000 \times n}{250}}$

$\longmapsto \large{\tt 1.5 = \cancel{\dfrac{1000 \times n}{250}}}$

$\longmapsto \large{\tt 1.5 = 4 \times n}$

$\longmapsto \large{\underline{\underline{\tt n = \dfrac{1.5}{4}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From the Formula we Know,

$\large \bigstar \; {\boxed{\tt n = \dfrac{w}{M}}}$

$\bold{Here}\begin{cases}\text{n Denotes Number of moles} \\ \text{W Denotes Weight} \\ \text{M Denotes Molar mass}\end{cases}$

Now,

$\large{\boxed{\tt n = \dfrac{w}{M}}}$

Substituting the Values,

$\longmapsto \large{\tt \dfrac{1.5}{4} = \dfrac{W}{40} \; \; \because \Big[\textsf{Molar mass _{\sf (NaOH)} = 40 \;g} \Big]}$

$\longmapsto \large{\tt \dfrac{1.5}{4} \times 40 = W}$

$\longmapsto \large{\tt W = \dfrac{1.5}{4} \times 40 }$

$\longmapsto \large{\tt W = \dfrac{1.5 \times 40}{4}}$

$\longmapsto \large{\tt W = \cancel{\dfrac{1.5 \times 40}{4}}}$

$\longmapsto \large{\tt W = 1.5 \times 10}$

$\longmapsto\large{\underline{\boxed{\red{\tt W = 15 \; grams}}}}$

∴ The weight of the NaOH will be 15 grams.

$\rule{300}{1.5}$

Answer 2

$\huge\bold\green{Question}$

How many grams of NaOH will be need to make250ml of 1.5M solution

$\huge\bold\green{Answer}$

$\begin{lgathered}\bold\green{Given}\begin{cases}\text{Vol. of solution = 250 ml.} \\ \text{Molarity of solution = 1.5 M }\end{cases}\end{lgathered}$

Hence , we have the molarity of the solution . So we know that

$\tt\blue\implies{ M = \dfrac{n}{V_{(In \; ml)}} \times 1000}$

• n = Number of moles
• V = Denotes Volume (In ml)
• M = Denotes Molarity}

Hence ,

$\tt\implies{ M = \dfrac{n}{V_{(In \; ml)}} \times 1000}$

So , simply by subsituting the known values in formula we get

= $\sf{ 1.5 \; M = \dfrac{n}{250 \; ml} \times 1000}$

= $\sf{ 1.5 = \dfrac{n}{250} \times 1000}$

= $\sf{ 1.5 = \dfrac{1000 \times n}{250}}$

= $\sf{ 1.5 = \cancel{\dfrac{1000 \times n}{250}}}$

= $\sf\blue{ 1.5 = 4 \times n}$

Hence , from the above data and given formula we can conclude that :-

$\sf\blue\implies{ n = \dfrac{w}{M}}$

• n = Number of moles
• W = Weight
• M = Molar mass

So , by using this formula and by subsituting the known values in formula we get :-

$\sf{ \dfrac{1.5}{4} = \dfrac{W}{40} \; \; \because \Big[\textsf{Molar mass _{\sf (NaOH)} = 40 \;g} \Big]}$

= $\sf{\dfrac{1.5}{4} \times 40 = W}$

= $\sf{ W = \dfrac{1.5}{4} \times 40 }$

= $\sf{ W = \dfrac{1.5 \times 40}{4}}$

= $\sf{ W = \dfrac{1.5 \times 40}{4}}$

= $\sf{ W = 1.5 \times 10}$

Hence , the accurate weight of NaOH is 15 grams