Question #28b04

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Question #28b04

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Answer 1 · 2017-12-13T15:17:54

$0.01875 m o l l d m^{- 3}$ $0.01875molcolor(white)(l)dm^(-3)$

Explanation:

$1.5 m g = 0.0015 g$ $1.5mg=0.0015g$

$n (N a O H) = \frac{0.0015}{40} = 3.75 \cdot 10^{- 5} m o l$ $n(NaOH)=0.0015/40=3.75*10^(-5)mol$

$c_{1} v_{1} = c_{2} v_{2}$ $c_1v_1=c_2v_2$ , where $c$ $c$ is concentration and $v$ $v$ is volume

$25 m L = 0.025 L = 0.025 d m^{3}$ $25mL=0.025L=0.025dm^3$

$[N a O H] = \frac{3.75 \cdot 10^{- 5}}{0.025} = 0.15 m o l l^{d} m (- 3)$ $[NaOH]=(3.75*10^(-5))/0.025=0.15molcolor(white)(l)^dm(-3)$

$c_{2} = \frac{c_{1} v_{1}}{v_{2}} = \frac{0.15 \cdot 0.025}{0.250} = 0.015 m o l l d m^{- 3}$ $c_2=(c_1v_1)/v_2=(0.15*0.025)/0.250=0.015molcolor(white)(l)dm^(-3)$

$\frac{c_{a} v_{a}}{n_{a}} = \frac{c_{b} v_{b}}{n_{b}}$ $(c_av_a)/(n_a)=(c_bv_b)/(n_b)$

As the mole ratio of NaOH:HCl is 1:1, $n_{a} = n_{b} = 1$ $n_a=n_b=1$

$c_{a} v_{a} = c_{b} v_{b}$ $c_av_a=c_bv_b$

$c_{a} = \frac{c_{b} v_{b}}{v_{a}} = \frac{0.025 (0.015)}{0.020} = 0.01875 m o l l d m^{- 3}$ $c_a=(c_bv_b)/(v_a)=(0.025(0.015))/(0.020)=0.01875molcolor(white)(l)dm^(-3)$

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Question #28b04

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