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Câu hỏi: Hòa tan 4,68 gam kali vào 50 gam nước. Nồng độ phần trăm của dung dịch thu được là
A. $8,58 \%$.
B. $8,56 \%$.
C. $12,32 \%$.
D. $12,29 \%$.
A. $8,58 \%$.
B. $8,56 \%$.
C. $12,32 \%$.
D. $12,29 \%$.
$2 \mathrm{~K}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_{2}$
${{n}_{\text{KOH}}}={{n}_{\text{K}}}=\dfrac{4,68}{39}=0,12~\text{mol}$
${{n}_{{{\text{H}}_{2}}}}=\dfrac{1}{2}\cdot {{n}_{\text{K}}}=0,06~ \text{mol}$
$m_{\mathrm{dd}}=m_{\mathrm{KL}}+m_{\mathrm{H}_{2} \mathrm{O}}-m_{H_{2}}=4,68+50-0,06.2=54,56$ gam
$C\%=\dfrac{{{m}_{\text{ct}}}}{{{m}_{\text{dd}}}}.100=\dfrac{4,68}{54,56}.100=8,56\%$.
${{n}_{\text{KOH}}}={{n}_{\text{K}}}=\dfrac{4,68}{39}=0,12~\text{mol}$
${{n}_{{{\text{H}}_{2}}}}=\dfrac{1}{2}\cdot {{n}_{\text{K}}}=0,06~ \text{mol}$
$m_{\mathrm{dd}}=m_{\mathrm{KL}}+m_{\mathrm{H}_{2} \mathrm{O}}-m_{H_{2}}=4,68+50-0,06.2=54,56$ gam
$C\%=\dfrac{{{m}_{\text{ct}}}}{{{m}_{\text{dd}}}}.100=\dfrac{4,68}{54,56}.100=8,56\%$.
Đáp án B.