How would you prepare a 1.0L solution of MgCl2 with a concentration of 1.4 M
Answers
Answer:
Given that , \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x}
p
1
+
q
1
+
x
1
=
p+q+x
1
.
Need To Find : Value of x ?
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀Given that ,
\begin{gathered}\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:\\\\ \end{gathered}
⇢
p
1
+
q
1
+
x
1
=
p+q+x
1
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Solving \: the \: Given \::}}\\\end{gathered}
⋆NowBySolvingtheGiven:
\begin{gathered}\dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x} \:\\\\ \dashrightarrow \sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} - \dfrac{1}{p + q + x} = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{1}{p} + \dfrac{1}{q}\bigg] +\bigg[ \dfrac{1}{x} - \dfrac{1}{p + q + x} \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{(x + p + q )- x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{x + p + q - x }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{p + q }{pq}\bigg] +\bigg[ \dfrac{ p + q }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{1}{pq} + \dfrac{ 1 }{x ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xp + xq + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x^2 + xq + xp + pq }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ x( x + q ) + p ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\\dashrightarrow \sf ( p + q ) \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = \dfrac{0}{( p + q )} \:\\\\ \dashrightarrow \sf \bigg[ \dfrac{ ( x + p ) ( x + q ) }{ (pq) (x) ( p + q + x ) } \bigg] = 0 \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \times (pq) (x) ( p + q + x ) \:\\\\ \dashrightarrow \sf ( x + p ) ( x + q ) = 0 \:\\\\ \dashrightarrow \sf \:x\:=\:-p \:\:or\:\: x \:=\:-q \:\\\\ \end{gathered}
⇢
p
1
+
q
1
+
x
1
=
p+q+x
1
⇢
p
1
+
q
1
+
x
1
−
p+q+x
1
=0
⇢[
p
1
+
q
1
]+[
x
1
−
p+q+x
1
]=0
⇢[
pq
p+q
]+[
x(p+q+x)
(x+p+q)−x
]=0
⇢[
pq
p+q
]+[
x(p+q+x)
x+p+q−x
]=0
⇢[
pq
p+q
]+[
x(p+q+x)
x+p+q−x
]=0
⇢[
pq
p+q
]+[
x(p+q+x)
p+q
]=0
⇢(p+q)[
pq
1
+
x(p+q+x)
1
]=0
⇢(p+q)[
(pq)(x)(p+q+x)
x
2
+xp+xq+pq
]=0
⇢(p+q)[
(pq)(x)(p+q+x)
x
2
+xq+xp+pq
]=0
⇢(p+q)[
(pq)(x)(p+q+x)
x(x+q)+p(x+q)
]=0
⇢(p+q)[
(pq)(x)(p+q+x)
(x+p)(x+q)
]=0
⇢[
(pq)(x)(p+q+x)
(x+p)(x+q)
]=
(p+q)
0
⇢[
(pq)(x)(p+q+x)
(x+p)(x+q)
]=0
⇢(x+p)(x+q)=0×(pq)(x)(p+q+x)
⇢(x+p)(x+q)=0
⇢x=−porx=−q
\begin{gathered}\dashrightarrow \:\underline {\boxed{\purple {\pmb{\frak{\:\:x\:=\:-p \:\:or\:\: \:-q \:}}}}}\:\:\bigstar \:\:\\\\\end{gathered}
⇢
x=−por−q
x=−por−q
★
\begin{gathered}\qquad \therefore \underline {\sf Hence, The \:value \:of \: x \:can \: be \:\pmb{\bf{ - p \: or \:-q \:}}.}\\\end{gathered}
∴
Hence,Thevalueofxcanbe
−por−q
−por−q.
2)
Question:-
If ( 2 - √5 )/( 2 + √5 ) =a√5 + b , then find the value of a and b.
To Find:-
Find the value of a and b.
Solution:-
\dashrightarrow\sf \: \dfrac { 2 - \sqrt { 5 } } { 2 + \sqrt { 5 } } = a \sqrt { 5 } + b⇢
2+
5
2−
5
=a
5
+b
Rationalise the denominator
\dashrightarrow\sf \: \dfrac { 2 - \sqrt { 5 } } { 2 + \sqrt { 5 } } \times \dfrac { 2 - \sqrt { 5 } } { 2 - \sqrt { 5 } } = a \sqrt { 5 } + b⇢
2+
5
2−
5
×
2−
5
2−
5
=a
5
+b
\dashrightarrow\sf \: \dfrac { { ( 2 - \sqrt { 5 } ) }^{ 2 } } { ( 2 + \sqrt { 5 } )( 2 - \sqrt { 5 } ) } = a \sqrt { 5 } + b⇢
(2+
5
)(2−
5
)
(2−
5
)
2
=a
5
+b
\dashrightarrow\sf \: \dfrac { 4 + 5 - 4 \sqrt { 5 } } { 4 - 5 } = a \sqrt { 5 } + b⇢
4−5
4+5−4
5
=a
5
+b
\dashrightarrow\sf \: \dfrac { - ( 9 + 4 \sqrt { 5 } ) } { - 1 } = a \sqrt { 5 } + b⇢
−1
−(9+4
5
)
=a
5
+b
\dashrightarrow\sf \: 9 + 4 \sqrt { 5 } = a \sqrt { 5 } + b⇢9+4
5
=a
5
+b
Hence ,
a = 4 , b = 9
Explanation:
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