\(m_{ddBaCl_2}=15+200=215\left(g\right)\)

\(C\%=\dfrac{15}{215}.100\%\approx6,98\%\)

\(C\%_{ddNaOH\left(thu.được\right)}=\dfrac{20}{20+150}.100\%\approx11,765\%\)

* Ý 1 :

m_{dd sau trộn} = m_ct + m_H2O = 50 + 250 = 300(g)

C%_{dd thu được} = \(\dfrac{m_{ct}}{m_{dd}}.100\%=\dfrac{50}{300}.100\%=16,67\%\)

*Ý 2 : Đổi 250ml = 0,25(l)

n_BaCl2 = 18_/208 =9_/104 (mol)

=> C_M của dd BaCl₂ = n_/V = 9_/104 : 0,25 =0,35(M)

a. m_dd = 50 + 250 = 300g

Nồng độ phần trăm dung dịch thu được là:

C%= m_ct / m_dd = 50 / 300 x 100 = 16,7%

b. nBaCl₂= m / M = 18 / 208 = 0,09 mol

Nồng độ mol dung dịch BaCl₂ là:

C_M= n/ V = 0,09/ 250 = 0,00036 M

Chúc bạn học tốt!

\(a.\)

\(m_{dd}=10+40=50\left(g\right)\)

\(C\%=\dfrac{10}{50}\cdot100\%=20\%\)

\(b.\)

\(m_{KOH}=0.25\cdot56=14\left(g\right)\)

\(m_{dd_{KOH}}=14+36=50\left(g\right)\)

\(C\%_{KOH}=\dfrac{14}{50}\cdot100\%=28\%\)

Sửa đề: 9,2 gam Na

\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            0,4------------------>0,8

\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,4----------------->0,8

\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)

a, \(C\%_{NaOH}=\dfrac{4}{4+2,8+118,2}.100\%=3,2\%\)

\(C\%_{KOH}=\dfrac{2,8}{4+2,8+118,2}.100\%=2,24\%\)

b, \(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,1}{0,125}=0,8\left(M\right)\)

\(n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(\Rightarrow C_{M_{KOH}}=\dfrac{0,05}{0,125}=0,4\left(M\right)\)

\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

\(0.1............................0.2\)

\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)

\(m_{dd_{H_3PO_4}}=14.2+185.8=200\left(g\right)\)

\(C\%H_3PO_4=\dfrac{19.6}{200}\cdot100\%=9.8\%\)

\(V_{dd_{H_3PO_4}}=\dfrac{200}{1.1}=181.8\left(ml\right)=0.1818\left(l\right)\)

\(C_{M_{H_3PO_4}}=\dfrac{0.2}{0.1818}=1.1\left(M\right)\)

Ta có: \(n_{P_2O_5}=\dfrac{14,2}{142}=0,1\left(mol\right)\)

PT: \(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

____0,1_____________0,2 (mol)

\(\Rightarrow m_{H_3PO_4}=0,2.98=19,6\left(g\right)\)

Có: m _{dd sau pư} = m_P2O5 + m_H2O = 200 (g)

\(\Rightarrow C\%_{H_3PO_4}=\dfrac{19,6}{200}.100\%=9,8\%\)

Có: V _{dd sau pư} = \(\dfrac{200}{1,1}=\dfrac{2000}{11}\left(ml\right)=\dfrac{2}{11}\left(l\right)\)

\(\Rightarrow C_{M_{H_3PO_4}}=\dfrac{0,2}{\dfrac{2}{11}}=1,1M\)

Bạn tham khảo nhé!

a) Ta có: \(C\%_{NaCl}=\dfrac{60}{60+1250}\cdot100\%\approx4,58\%\)

b) Ta có: \(C_{M_{Na_2CO_3}}=\dfrac{0,5}{0,5}=1\left(M\right)\)