5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL ) of this solution would be required to prepare 500 mL of 0.1 M solution? Given: Molar Mass of Na, O and H is 23,16 and 1 g mol -1 respectively

Question

5 g of NaOH was dissolved in deionized water to prepare a 450 mL stock solution. What volume (in mL ) of this solution would be required to prepare 500 mL of 0.1 M solution? Given: Molar Mass of Na, O and H is 23,16 and 1 g mol -1 respectively (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

M=(5/40) × (1000/450) M1 V1=M2 V2 ((5/40) × (1000/450)) × V1=0.1 × 500 V1=180

Q. $5 g$ of $N a O H$ was dissolved in deionized water to prepare a $450 m L$ stock solution. What volume (in $m L$ ) of this solution would be required to prepare $500 m L$ of $0.1 M$ solution?
Given: Molar Mass of $N a, O$ and $H$ is 23,16 and $1 g m o l^{- 1}$ respectively

$M = \frac{5}{40} \times \frac{1000}{450}$
$M_{1} V_{1} = M_{2} V_{2}$
$(\frac{5}{40} \times \frac{1000}{450}) \times V_{1} = 0.1 \times 500$
$V_{1} = 180$

Q. 5g of NaOH was dissolved in deionized water to prepare a 450mL stock solution. What volume (in mL ) of this solution would be required to prepare 500mL of 0.1M solution? Given: Molar Mass of Na,O and H is 23,16 and 1gmol−1 respectively

M=405​×4501000​ M1​V1​=M2​V2​ (405​×4501000​)×V1​=0.1×500 V1​=180

Q. $5 g$ of $N a O H$ was dissolved in deionized water to prepare a $450 m L$ stock solution. What volume (in $m L$ ) of this solution would be required to prepare $500 m L$ of $0.1 M$ solution?
Given: Molar Mass of $N a, O$ and $H$ is 23,16 and $1 g m o l^{- 1}$ respectively

$M = \frac{5}{40} \times \frac{1000}{450}$
$M_{1} V_{1} = M_{2} V_{2}$
$(\frac{5}{40} \times \frac{1000}{450}) \times V_{1} = 0.1 \times 500$
$V_{1} = 180$