1) Ở 10 độ C, hoà tan hoàn toàn m gam KCl vào 200g dd KCl 10% thu được dd bão hoà. Tính m biết độ tan của KCl ở 10 độ C là 31,2g 2) Ở 10 độ C, dd K2Cr2O7 bão hoà có nồng độ 6,54% . Tính độ tan của K2Cr2O7 ở 10 độ C và khối lượng K2Cr2O7 có trong 200g dd bão hoà 3) Ở 10 độ C, khi

tlh140405

2 năm trước

1) Ở 10 độ C, hoà tan hoàn toàn m gam KCl vào 200g dd KCl 10% thu được dd bão hoà. Tính m biết độ tan của KCl ở 10 độ C là 31,2g 2) Ở 10 độ C, dd K2Cr2O7 bão hoà có nồng độ 6,54% . Tính độ tan của K2Cr2O7 ở 10 độ C và khối lượng K2Cr2O7 có trong 200g dd bão hoà 3) Ở 10 độ C, khi hoà tan hoàn toàn m gam KNO3 vào a gam dd KNO3 10% thu được 200g dd KNO3 bão hoà . Biết độ tan của KNO3 ở 10 độ C là 21,9g. Tính a,m

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manhh4147

$\begin{array}{l}
1)\\
mKCl\,trong\,{\rm{dd}}\,10\% = 200 \times 10\% = 20g\\
m{H_2}O = 200 - 20 = 180g\\
mKCl\,trong\,{\rm{dd}}bh = \dfrac{{180 \times 31,2}}{{100}} = 56,16g\\
m = 56,16 - 20 = 36,16g\\
2)\\
m{K_2}C{r_2}{O_7} = \dfrac{{200 \times 6,54}}{{100}} = 13,08g\\
m{H_2}O = 200 - 13,08 = 186,92g\\
S = \dfrac{{13,08}}{{186,92}} \times 100 = 7g\\
3)
\end{array}$

$\begin{array}{l}
mKN{O_3}\,trong\,{\rm{dd}}\,10\% = 10\% \times a = 0,1a\,(g)\\
m{H_2}O = a - 0,1a = 0,9a(g)\\
mKN{O_3}\,trong\,{\rm{dd}}\,bh = \dfrac{{0,9a \times 21,9}}{{100}} = 0,1971a(g)\\
a + m = 200(1)\\
0,1a + m = 0,1971a(2)\\
(1) - (2) \Rightarrow 0,9a = 200 - 0,1971a \Rightarrow a = 182,3g\\
m = 200 - 182,3 = 17,7g
\end{array}$

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motsombungnangha

Đáp án:

Giải thích các bước giải:

$\begin{array}{l} 1) \\ m K C l t r o n g d d 10 % = 200 \times 10 % = 20 g \\ m H_{2} O = 200 - 20 = 180 g \\ m K C l t r o n g d d b h = \frac{180 \times 31, 2}{100} = 56, 16 g \\ m = 56, 16 - 20 = 36, 16 g \\ 2) \\ m K_{2} C r_{2} O_{7} = \frac{200 \times 6, 54}{100} = 13, 08 g \\ m H_{2} O = 200 - 13, 08 = 186, 92 g \\ S = \frac{13, 08}{186, 92} \times 100 = 7 g \\ 3) \end{array}$

$\begin{array}{l} m K N O_{3} t r o n g d d 10 % = 10 % \times a = 0, 1 a (g) \\ m H_{2} O = a - 0, 1 a = 0, 9 a (g) \\ m K N O_{3} t r o n g d d b h = \frac{0, 9 a \times 21, 9}{100} = 0, 1971 a (g) \\ a + m = 200 (1) \\ 0, 1 a + m = 0, 1971 a (2) \\ (1) - (2) \Rightarrow 0, 9 a = 200 - 0, 1971 a \Rightarrow a = 182, 3 g \\ m = 200 - 182, 3 = 17, 7 g \end{array}$

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